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questions</a></label></li></ul></div> <button id="show-periodic-table" class="btn btn-default btn-block" style="margin-bottom: 10px"><i class="fa fa-table"></i> Periodic table</button> <div class="hidden-xs hidden-sm"> <button class="btn btn-default btn-block text-xs-center" data-toggle="modal" data-target="#modal-feedback" style="margin-bottom: 10px"><i class="fa fa-send"></i> Feedback</button> </div> </div> <div class="col-md-9" id="main-column"> <h1 class="page_title"> Atomic Structure AHL (HL only) paper 1 questions <a href="#" class="mark-page-favorite pull-right" data-pid="2125" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../chemistry.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../2075/paper-1-exam-questions.html">Paper 1 Exam Questions</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray"> Atomic Structure AHL (HL only) paper 1 questions</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 30 minutes"><i class="fa fa-clock-o"></i> 30'</span> </ol> <article id="main-article"> <p><strong>Topic 12.1</strong></p> <p><strong>Paper 1 style questions </strong>are multiple choice. You are <strong>not permitted to use a calculator or the data book</strong> for these questions, but you should use a periodic table.</p> <p>A <strong>periodic table pop-up</strong> is available on the left hand menu.</p> <div class="greenBg"> <div class="tib-quiz" data-stats="7-579-2125"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which element is in group 13?</p><p><img alt="" src="../../images/atomic-structure/screenshot-2021-05-02-at-10.34.48.png" style="width: 600px; height: 137px;"></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> X</label></p><p><label class="radio"> <input type="radio"> Z</label></p><p><label class="radio"> <input type="radio"> <span>Y</span></label> </p><p><label class="radio"> <input type="radio"> <span>W</span></label> </p></div><div class="q-explanation"><p>Group 13 elements (like B or Al) have the largest jump in ionisation energies (IEs) between the removal of the third and fourth electrons, because the first three electrons are removed from a p sub-level and s sub-level that are all within the same energy level (The electron configuration of Boron for example is 1s<sup>2</sup> 2s<sup>2</sup> 2p<sup>1</sup>). The fourth electron, however, is removed from a new energy level which is lower in energy, closer to the nucleus with stronger nuclear attraction and less shielding making the electrons more difficult to remove.</p><p>Element X has gradually increasing 1st to 3rd IEs and then a dramatic jump (much greater in comparison to the differences between the first three IEs) in IE to the 4th, so element X is most likely to be in group 13.</p><p><strong>X</strong> is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which graph shows the trend in first ionisation energies of elements across period 2?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-10.10.54.png" style="width: 200px; height: 232px;"> </label></p><p><label class="radio"> <input type="radio"> <img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-10.11.05.png" style="width: 200px; height: 229px;"></label></p><p><label class="radio"> <input type="radio"> <img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-10.11.37.png" style="width: 200px; height: 228px;"></label></p><p><label class="radio"> <input type="radio"> <img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-10.05.19.png" style="width: 200px; height: 230px;"></label></p></div><div class="q-explanation"><p>The graph showing the trend in first ionisation energies (1st IE) across the periodic table needs to be learned (at least up to calcium, atomic number 20) such that you are able to sketch the graph. The graph is shown below:</p><p><img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-10.23.13.png" style="width: 600px; height: 341px;"></p><p>There are only two elements in the first period, H and He.</p><p>The best way to 'anchor' the position of a graph showing a section of the trend in ionisation energies is to look for the highest points which are always in group 18 (Noble gases) and the lowest points which are always group 1 (Alkali metals). Also, notice the similar pattern across the eight elements in period 2 and period 3. There are regular dips in 1st IE in the pattern 2, 3, 3 across both periods, before a big drop to the next period. The pattern is explained in the video: <a href="../716/electrons-in-atoms.html"><strong>First ionisation energies</strong></a> </p><p>The 2, 3, 3 pattern is therefore seen across period 2 in the correct answer:</p><p><img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-10.10.54.png" style="width: 200px; height: 232px;"></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">3</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which element is represented by the first eight successive ionisation energies shown on the graph?</p><p><img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-12.15.13.png" style="width: 600px; height: 313px;"></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> P </label></p><p><label class="radio"> <input type="radio"> Ar </label></p><p><label class="radio"> <input type="radio"> Si</label></p><p><label class="radio"> <input type="radio"> Al</label></p></div><div class="q-explanation"><p>The electron configurations for the four possible elements are; aluminium 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>1</sup> ; silicon 1s<sup>2</sup> 2s<sup>2</sup> 2p<sup>6</sup> 3s<sup>2</sup> 3p<sup>2</sup> ; phosphorus 1s<sup>2</sup> 2s<sup>2</sup> 2p<sup>6</sup> 3s<sup>2</sup> 3p<sup>3</sup> and argon 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>6</sup>.</p><p>The graph has the largest jump in ionisation energy (IE) between the 5th and 6th electron and a relatively smaller jump between the 3rd and 4th electron. The large jump in IE (5th to 6th) will correspond to a new energy level, and the smaller jump to a new sub-level.</p><p>Phosphorus, P, is the only element in the list with a change in energy level after the removal of five electrons (3p<sup>3</sup> and 3s<sup>2</sup> removed) and it also has a change in sub-level (3p to 3s) after three electrons have been removed.</p><p><strong>P</strong> is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>The first four successive ionisation energies for two elements, X and Z, are given below (in kJ mol<sup>−1</sup>).</p><p>Which pair of elements represent X and Z?.</p><p><strong>X:</strong> 1314, 3388, 5301, 7469</p><p><strong>Z: </strong> 787, 1577, 3232, 4356</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>X is carbon; Z is sulfur</span></label></p><p><label class="radio"> <input class="c" type="radio"> <span>X is oxygen; Z is silicon</span></label> </p><p><label class="radio"> <input type="radio"> <span>X is carbon; Z is silicon</span></label> </p><p><label class="radio"> <input type="radio"> X is oxygen; Z is sulfur<span></span></label> </p></div><div class="q-explanation"><p>The electron configurations for the four possible elements are; oxygen 1s<sup>2</sup> 2s<sup>2</sup> 2p<sup>4</sup> ; silicon 1s<sup>2</sup> 2s<sup>2</sup> 2p<sup>6</sup> 3s<sup>2</sup> 3p<sup>2</sup> ; carbon 1s<sup>2</sup> 2s<sup>2</sup> 2p<sup>2</sup> and sulfur 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>4</sup>.</p><p>Element X has a similar difference of approx. 2000 units between each ionisation energy (IE), with no larger jumps in IE. For oxygen there will be no larger jumps in ionisation energy, since the first four electrons are all removed from the 2p sub-level, so element X is oxygen. (Carbon would show a slightly larger jump is IE between the second and third IE; 2p to 2s.)</p><p>Element Z has the largest difference of approx. 1700 units between the second and third IEs. For silicon the largest jump will be between the second and third IEs (3p to 3s), so element Z is silicon. (Sulfur would have a similar difference between all of the first four IEs as all electrons are removed from 3p.)</p><p>The correct answer is therefore <strong>X is oxygen; Z is silicon</strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">5</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which transition on the diagram could represent the first ionisation energy of hydrogen?</p><p><img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-14.23.06.png" style="width: 400px; height: 278px;"></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>A</span></label> </p><p><label class="radio"> <input type="radio"> <span>D</span></label> </p><p><label class="radio"> <input type="radio"> <span>B</span></label> </p><p><label class="radio"> <input type="radio"> <span>C</span></label> </p></div><div class="q-explanation"><p>A hydrogen atom has only one electron. In the ground state, the electron is in the 1s sub-level (in the n=1 energy level). The first ionisation energy involves removing an electron from every atom in one mole of gaseous atoms, this can be represented as moving the ground state electron to the n=∞ energy level.</p><p>Therefore an electron transition from n=1 to n=∞ could represent the first ionisation energy of hydrogen.</p><p><strong>A</strong> is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>The graph shows the first ionisation energies of some consecutive elements in order of atomic number. Which statement is correct?</p><p><img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-15.06.30.png" style="width: 300px; height: 393px;"></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>Element X is in group 18</span></label> </p><p><label class="radio"> <input type="radio"> <span>Element X is in group 2</span></label> </p><p><label class="radio"> <input type="radio"> <span>Element Z is group 5 </span></label> </p><p><label class="radio"> <input type="radio"> <span>Element Z is in group 2</span></label> </p></div><div class="q-explanation"><p>This graph of ionisation energy does not begin with hydrogen. The best way to 'anchor' the position of a graph showing a section of the trend in ionisation energies is to look for the highest points which are always in group 18 (Noble gases) and the lowest points which are always group 1 (Alkali metals). Look also for the 2, 3, 3 pattern in elements that is seen across period 2 and period 3. See the graph below:</p><p><img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-10.23.13.png" style="width:600px;height:341px;"></p><p>The graph showing the trend in first ionisation energies (1st IE) across the periodic table needs to be learned (at least up to calcium, atomic number 20) such that you are able to sketch the graph. The pattern is explained in the video: <a href="../716/electrons-in-atoms.html"><strong>First ionisation energies</strong></a> </p><p>In this question element X is at a high point on the graph that corresponds to group 18. Element Z is in group 13.</p><p>The correct answer is therefore <strong>Element X is in group 18</strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">7</div><div class="exercise shadow-bottom"><div class="q-question"><p>The first eight successive ionisation energies for an element are shown on the graph. Why is there a relatively large increase in successive ionisation energy between the 4th and 5th electrons removed?</p><p><img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-15.40.32.png" style="width: 600px; height: 317px;"></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> The 5th electron removed is in a lower sub-level</label></p><p><label class="radio"> <input class="c" type="radio"> The 5th electron removed is in a lower energy level </label></p><p><label class="radio"> <input type="radio"> The 4th electron removed is paired-up with another electron in the same orbital</label></p><p><label class="radio"> <input type="radio"> The 5th electron removed occupies the orbital singularly</label></p></div><div class="q-explanation"><p>The graph has the largest jump in ionisation energy (IE) between the 4th and 5th electrons and a relatively smaller jump between the 2nd and 3rd electrons.</p><p>Large jumps in successive IEs will always correspond to new (lower) energy levels, and the smaller jumps will often correspond to a new (lower) sub-levels.</p><p><strong>The 5th electron removed is in a lower energy level </strong>is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>The graph shows the first ionisation energies of the first twenty consecutive elements in order of atomic number. Which statement correctly explains a decrease in ionisation energy?</p><p><img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-16.03.05.png" style="width: 600px; height: 299px;"></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Element 13 has a higher nuclear charge than element 12</span></label></p><p><label class="radio"> <input class="c" type="radio"> <span>Element 16 has greater electron-electron repulsion than element 15 due to electron pairing</span></label></p><p><label class="radio"> <input type="radio"> <span>Element 3 has its outermost electron in a lower energy level than element 2</span></label></p><p><label class="radio"> <input type="radio"> A new sub-level is being filled in element 8 compared to element 7<span> </span></label> </p></div><div class="q-explanation"><p>The graph showing the trend in first ionisation energies (1st IE) across the periodic table needs to be learned (at least up to calcium, atomic number 20) such that you are able to sketch the graph. The pattern is explained in the video: <a href="../716/electrons-in-atoms.html"><strong>First ionisation energies</strong></a> </p><p>The large decreases across the graph are due to an electron being removed from a new energy level of <strong>higher energy </strong>e.g. element 2 is Helium (1s<sup>2</sup>) to element 3, which is Lithium (1s<sup>2</sup> 2s<sup>1</sup>).</p><p>The smaller decreases between elements 4 to 5 and 12 to 13 are due to an electron being removed from a new sub-level of higher energy e.g. element 4 is Berylium (1s<sup>2</sup>2s<sup>2</sup>) to element 5, which is Boron (1s<sup>2</sup> 2s<sup>2</sup> 2p<sup>1</sup>).</p><p>The decreases in IE between elements 7 to 8 (N to O), and 15 to 16 (P to S) are due to the electron being removed from a p orbital in which the electrons have been required to pair-up, creating additional electron-electron repulsion (relative to the previous three elements in which the electron removed occupies the p orbital singularly). </p><p>The correct answer is therefore <strong>Element 16 has greater electron-electron repulsion than element 15 due to electron pairing</strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">9</div><div class="exercise shadow-bottom"><div class="q-question"><p>A period 2 element, X, reacts with oxygen to form an ionic oxide with empirical formula X<sub>2</sub>O.</p><p>Which correctly represents the first three successive ionisation energies (in kJ mol<sup>−1</sup>) for element X?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> 1087, 2353, 4621<span></span></label> </p><p><label class="radio"> <input type="radio"> 900, 1757, 14849</label></p><p><label class="radio"> <input class="c" type="radio"> <span>520, 7298, 11815</span></label></p><p><label class="radio"> <input type="radio"> <span>1402, 2856, 4578</span></label> </p></div><div class="q-explanation"><p>The element X forms an ionic compound with oxygen, so must be a metal (ionic compounds are always metal and non-metal) and has formula X<sub>2</sub>O, so X forms a 1+ ion. X is in period 2, so it must be Lithium.</p><p>Lithium has electron configuration 1s<sup>2</sup> 2s<sup>1</sup>, so there will be a large relative increase in ionisation energy (IE) between the 1st and 2nd IEs, as the second electron is removed from a new energy level.</p><p>The only series that shows a large relative increase in IE between 1st and 2nd IE is <strong>520, 7298, 11815. </strong>Therefore,<strong> </strong>this is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">10</div><div class="exercise shadow-bottom"><div class="q-question"><p>The first ionisation energies (in kJ mol<sup>−1</sup>) of five <strong>consecutive</strong> elements on the periodic table are:</p><p>1251, 1521, 419, 590, 631.</p><p>What could these elements be?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> Elements in the d-block</label></p><p><label class="radio"> <input type="radio"> <span>The last three elements of one period and the first two elements of the next period</span></label> </p><p><label class="radio"> <input class="c" type="radio"> The last two elements of one period and the first three elements of the next period</label></p><p><label class="radio"> <input type="radio"> The first five e<span>lements in a p sub-level </span></label> </p></div><div class="q-explanation"><p>Large decreases in first ionisation energy (like 1521 to 419) across the periodic table always indicate a new period. The graph showing the trend in first ionisation energies (1st IE) across the periodic table needs to be learned (at least up to calcium, atomic number 20) such that you are able to sketch the graph. The pattern is explained in the video: <a href="../716/electrons-in-atoms.html"><strong>First ionisation energies</strong></a> </p><p><img alt="" src="../../images/atomic-structure/screenshot-2021-05-07-at-10.23.13.png" style="width:600px;height:341px;"></p><p>In this question the 1st IEs 1251 and 1521 suggest the last two elements in a period, as there is a big drop to 419 suggesting a new period.</p><p>The correct answer is therefore <strong>The last two elements of one period and the first three elements of the next period</strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="totals"><span class="score"></span><button class="btn btn-success btn-block text-center check-total"><i class="fa fa-check-square-o"></i> Check</button></div></div><hr> </div> <p><strong>Paper 1 style questions </strong>are multiple choice. You are <strong>not permitted to use a calculator or the data book</strong> for these questions, but you should use a periodic table.</p> <p>A <strong>periodic table pop-up</strong> is available on the left hand menu.</p> <div class="greenBg"> <div class="tib-quiz" data-stats="7-578-2125"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following best represents the shape of an s-atomic orbital?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <img alt="" src="../../images/atomic-structure/screenshot-2021-04-30-at-13.00.32.png" style="width: 60px; height: 82px;"></label></p><p><label class="radio"> <input class="c" type="radio"> <img alt="" src="../../images/atomic-structure/screenshot-2021-04-30-at-13.00.21.png" style="width: 60px; height: 53px;"> </label></p><p><label class="radio"> <input type="radio"> <span><img alt="" src="../../images/atomic-structure/screenshot-2021-04-30-at-13.00.51.png" style="width: 120px; height: 91px;"></span></label> </p><p><label class="radio"> <input type="radio"> <span><img alt="" src="../../images/atomic-structure/screenshot-2021-04-30-at-13.00.42.png" style="width: 120px; height: 61px;"></span></label> </p></div><div class="q-explanation"><p>Orbitals are plots of electron position (90% probability) around the nucleus. They are three dimensional (not planar). An s<em>-</em>orbital is a sphere; a p<em>-</em>orbital is a three dimensional figure-of-eight.</p><p>So the answer here that best <span data-scayt-word="decribes" data-wsc-id="ko49sfts63j3rcm5a" data-wsc-lang="en_GB">decribes</span> as s<em>-</em>orbital is <img alt="" src="../../images/atomic-structure/screenshot-2021-04-30-at-13.00.21.png" style="width: 60px; height: 53px;">.</p><p>The red figure-of-eight image represents a p-orbital.</p><p>For HL students only: The elongated blue image represents a sigma molecular orbital (two orbitals overlapping end on) and the elongated red image represents a pi molecular orbital (two p-orbitals overlapping side on).</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the correct electronic configuration for an atom of copper?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>1s<sup>2 </sup> 2s<sup>2</sup> 2p<sup>6</sup> 3s<sup>2</sup> 3p<sup>6</sup> 4s<sup>1</sup> 3d<sup>10</sup></span></label></p><p><label class="radio"> <input type="radio"> [Ar] <span>4s<sup>2</sup> 4p<sup>9</sup></span></label><sup><sup> </sup></sup></p><p><label class="radio"> <input type="radio"> [Ar] <span>4s<sup>2</sup> 3d<sup>9</sup></span></label><sup> </sup></p><p><label class="radio"> <input type="radio"> <span>1s<sup>2 </sup> 2s<sup>2</sup> 2p<sup>6</sup> 3s<sup>2</sup> 3p<sup>6</sup> 4s<sup>1</sup> 4d<sup>10</sup></span></label><sup> </sup></p></div><div class="q-explanation"><p>The sub-levels of an atom are filled in the following order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s. Remember that 's' sub-levels hold a maximum of 2 electrons, 'p' hold 6, and 'd' hold 10.</p><p>A copper atom has 29 electrons in total (atomic number is 29), <strong>but</strong> the configuration is not what would be expected as the electrons in the 4s and 3d energy levels are 4s<sup>1</sup> 3d<sup>10</sup> instead of filling the sub-levels in the conventional manner (4s<sup>2</sup> 3d<sup>9</sup>). Chromium and copper have unusual configurations and need to be learned.</p><p>Remember that [Ar] is shorthand and indicates the electron configuration of argon: 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>6</sup>.</p><p>The correct answer is therefore <strong>1s<sup>2 </sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>6</sup>4s<sup>1</sup>3d<sup>10</sup></strong></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">3</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which is the correct number of protons, neutrons and electrons in the calcium ion <span class="math-tex">\(\begin{smallmatrix} 44\\ 20 \end{smallmatrix}\)</span>Ca<sup>2</sup><sup>+</sup>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> 20<span> protons, 24 neutrons, 20 electrons</span></label> </p><p><label class="radio"> <input type="radio"> 24<span> protons, 20 neutrons, 24 electrons</span></label> </p><p><label class="radio"> <input type="radio"> 24<span> protons, 44 neutrons, 22 electrons</span></label> </p><p><label class="radio"> <input class="c" type="radio"> 20<span> protons, 24 neutrons, 18 electrons </span></label> </p></div><div class="q-explanation"><p>The upper number is the <em>mass number</em>, which is equal to protons plus neutrons. The lower number is the <em>atomic number</em> which is the number of protons. The number of electrons is equal to the number of protons in a neutral atom, but this is 2+ positive ion, so it has two fewer electrons than protons.</p><p>So the species has <strong>20 protons</strong>, (44−20) <strong>24 neutrons</strong> and <strong>18 electrons</strong> (two fewer than protons).</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which statements are correct with regard to the absorption spectrum of hydrogen?</p><p>1: The lines are produced when electrons move from higher to lower energy levels</p><p>2: The lines converge at lower wavelengths</p><p>3: The lines in the visible region of the electromagnetic spectrum are seen as coloured lines</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>1 & 3 only</span></label> </p><p><label class="radio"> <input class="c" type="radio"> 2<span> only</span></label> </p><p><label class="radio"> <input type="radio"> None of the statements</label></p><p><label class="radio"> <input type="radio"> <span>1 & 2 only</span></label> </p></div><div class="q-explanation"><p>The line absorption spectrum of hydrogen is composed of discrete lines that represent particular frequencies. Lines are produced by electron transitions from <strong>lower to higher energy levels</strong> (for absorption; from higher to lower for emission).</p><p><strong>The lines converge at</strong> higher frequencies/energies which corresponds to<strong> lower wavelengths</strong>.</p><p>Any lines in the <strong>absorption</strong> spectrum are <strong>not </strong>seen as coloured lines, but as black lines (an absence of colour) as electromagnetic radiation is absorbed (not emitted).</p><p>Thus the answer is <strong>2 only</strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">5</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which statement about <sup>65</sup>Zn<sup>2+</sup> and <sup>65</sup>Cu<sup>+</sup> is correct?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> Both have the same number of neutrons and electrons</label></p><p><label class="radio"> <input class="c" type="radio"> Both have the same number of electrons</label></p><p><label class="radio"> <input type="radio"> Both have the same number of neutrons</label></p><p><label class="radio"> <input type="radio"> Both have the same number of protons</label></p></div><div class="q-explanation"><p>The upper number on the left is the <em>mass number</em>, which is equal to protons plus neutrons. The <em>atomic number</em> for zinc is 30 and for copper is 29; this is the number of protons, shown on the periodic table.</p><p>The number of electrons is equal to the number of protons in a neutral atom, but these species are positive ions so will have more protons than electrons.</p><p><sup>65</sup>Zn<sup>2+</sup> therefore has 30 protons, 65−30=35 neutrons, and 30−2=28 electrons (as it is a 2+ ion).</p><p><sup>65</sup>Cu<sup>+</sup> therefore has 29 protons, 65−29=36 neutrons, and 29−1=28 electrons (as it is a 1+ ion).</p><p>The correct answer is therefore <strong>Both have the same number of electrons.</strong></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>In which set do all the species contain more neutrons than electrons?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <sup>8</sup>Li, <sup>31</sup>P, <sup>15</sup>N<sup>3−</sup></label></p><p><label class="radio"> <input type="radio"> <sup>8</sup>Li<sup>+</sup>, <sup>31</sup>P<sup>3−</sup>, <sup>15</sup>N</label></p><p><label class="radio"> <input type="radio"> <sup>8</sup>Li, <sup>31</sup>P<sup>3−</sup>, <sup>15</sup>N<sup>3−</sup></label></p><p><label class="radio"> <input class="c" type="radio"> <sup>8</sup>Li<sup>+</sup>, <sup>31</sup>P, <sup>15</sup>N</label></p></div><div class="q-explanation"><p>The upper (superscript) number on the left is the <em>mass number</em>, which is equal to protons plus neutrons. The <em>atomic number</em>, which is the number of protons is shown on the periodic table. The number of electrons is equal to the number of protons in a neutral atom but negative ions will have correspondingly more electrons and positive ions correspondingly fewer.</p><p><sup>8</sup>Li<sup>+</sup> has 8−3=5 neutrons and 3−1=2 electrons</p><p><sup>8</sup>Li has 8−3=5 neutrons and 3 electrons</p><p><sup>31</sup>P<sup>3−</sup> has 31−15=16 neutrons and 15+3=18 electrons</p><p><sup>31</sup>P has 31−15=16 neutrons and 15 electrons</p><p><sup>15</sup>N<sup>3−</sup> has 15−7=8 neutrons and 7+3=10 electrons</p><p><sup>15</sup>N has 15−7=8 neutrons and 7 electrons</p><p>The correct answer is therefore <strong><sup>8</sup>Li<sup>+</sup>, <sup>31</sup>P, <sup>15</sup>N</strong> as this is the only set in which <u>all</u> species contain more neutrons than electrons.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">7</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which electron transition in the hydrogen emission spectrum emits radiation with the longest wavelength?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>n=3 to n=1</span></label></p><p><label class="radio"> <input type="radio"> n=1 to n=4</label></p><p><label class="radio"> <input type="radio"> <span>n=6 to n=7</span></label></p><p><label class="radio"> <input class="c" type="radio"> <span>n=6 to n=5</span></label></p></div><div class="q-explanation"><p>Longer wavelength corresponds to lower frequency/lower energy. So the longest wavelength will correspond to the electron transition of lowest energy.</p><p>The question also asks about emission - energy given out - so the transition must be from a <strong>higher to lower </strong>energy level.</p><p>The energy gaps between the levels get smaller at higher energy.</p><p>Therefore the smallest energy transition that is also an emission is <strong>n=6 to n=5</strong>, which is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>The full electron configuration of an atom of an element is 1s<sup>2</sup> 2s<sup>2</sup> 2p<sup>6</sup> 3s<sup>2</sup> 3p<sup>6</sup> 4s<sup>2</sup> 3d<sup>3</sup>.</p><p>To which group and period does the element belong?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Group 3, Period 4</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Group 5, Period 4</span></label></p><p><label class="radio"> <input type="radio"> <span>Group 3, Period 3</span></label> </p><p><label class="radio"> <input type="radio"> <span>Group 5, Period 3</span></label> </p></div><div class="q-explanation"><p>The electron configuration reflects an element's position on the periodic table.</p><p>The easiest way to solve this is to count the total number of electrons (23); as this is an atom, the number of electrons will be equal to the number of protons, so the element is vanadium.</p><p>Vanadium is in <strong>Group 5, Period 4</strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">9</div><div class="exercise shadow-bottom"><div class="q-question"><p>Fluorine has only one stable isotope. Which statement is incorrect<strong> </strong>for the fluoride ion <span class="math-tex">\(\begin{smallmatrix} 19\\ 9 \end{smallmatrix}\)</span>F<sup>−</sup>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> The ion contains the same number of electrons as neutrons.</label></p><p><label class="radio"> <input type="radio"> The nucleus of the ion contains 19 sub-atomic particles.</label></p><p><label class="radio"> <input type="radio"> All naturally occurring stable fluoride ions have 10 neutrons.</label></p><p><label class="radio"> <input class="c" type="radio"> The ion has the same electron configuration as an atom of argon. </label></p></div><div class="q-explanation"><p>The upper number is the <em>mass number</em>, which is equal to protons plus neutrons (<strong>19 sub-atomic particles in the nucleus</strong>). The lower number is the <em>atomic number</em> which is the number of protons. The number of electrons is equal to the number of protons in a neutral atom, but this is a 1− negative ion, so it has one more electron than protons.</p><p>So this fluoride ion has 9 protons, (19−9=)<strong>10 neutrons and 10 electrons </strong>(1s<sup>2</sup> 2s<sup>2</sup> 2p<sup>6</sup>).</p><p>The question states that there is only one stable isotope of fluorine so <strong>all stable fluoride ions will have 10 neutrons</strong>.</p><p>The correct answer is therefore <strong>The ion has the same electron configuration as an atom of argon </strong>as this is incorrect: Argon has an electron configuration of 1s<sup>2</sup> 2s<sup>2</sup> 2p<sup>6</sup> 3s<sup>2</sup> 3p<sup>6</sup>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">10</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the number of protons and the number of neutrons in <sup>82</sup>Kr?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> 46 protons, 36 neutrons</label></p><p><label class="radio"> <input class="c" type="radio"> 36 protons, 46 neutrons</label></p><p><label class="radio"> <input type="radio"> 36 protons, 82 neutrons</label></p><p><label class="radio"> <input type="radio"> 82 protons, 36 neutrons</label></p></div><div class="q-explanation"><p>The upper number on the left is the <em>mass number</em>, which is equal to protons plus neutrons. The <em>atomic number</em> for krypton is 36, which is the number of protons; shown on the periodic table.</p><p><sup>82</sup>Kr therefore has 36 protons and (82−36=)46 neutrons.</p><p>The correct answer is therefore <strong>36 protons, 46 neutrons</strong><strong>.</strong></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">11</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the correct condensed electronic configuration for the V<sup>3+</sup> ion?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> [Ar] 4s<sup>2</sup> 3d<sup>3</sup></label></p><p><label class="radio"> <input type="radio"> <span>1s<sup>2 </sup> 2s<sup>2</sup> 2p<sup>6</sup> 3s<sup>2</sup> 3p<sup>6</sup> 4s<sup>2</sup> 3d<sup>3</sup></span></label><sup><sup> </sup></sup></p><p><label class="radio"> <input class="c" type="radio"> [Ar] 4s<sup>0</sup> 3d<sup>2</sup></label></p><p><label class="radio"> <input type="radio"> [Ar] 4s<sup>2</sup> 3d<sup>0<sup> </sup></sup></label></p></div><div class="q-explanation"><p>The sub-levels of an atom are filled in the following order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s.</p><p>Remember that 's' sub-levels hold a maximum of 2 electrons, 'p' hold 6, and 'd' hold 10, and that [Ar] is shorthand and indicates the electron configuration of argon: 1s<sup>2</sup>2s<sup>2</sup>2p<sup>6</sup>3s<sup>2</sup>3p<sup>6</sup>.</p><p>A vanadium <strong>atom </strong>has 23 electrons in total (atomic number is 23) so the condensed <strong>atomic </strong>electron configuration is [Ar] 4s<sup>2</sup> 3d<sup>3</sup>.</p><p>The V<sup>3+</sup> ion has a 3+ charge so has lost three electrons. These electrons are lost from the 4s orbital <strong>before</strong> the 3d. Remember that '4s are gained first and lost first'! So two electrons are lost from 4s and one electron is lost from 3d.</p><p>The correct answer is therefore <strong>[Ar] 4s<sup>0</sup> 3d<sup>2</sup></strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">12</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the maximum number of electrons that can occupy the n=3 energy level?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> 8</label></p><p><label class="radio"> <input type="radio"> 14 </label></p><p><label class="radio"> <input type="radio"> 32 </label></p><p><label class="radio"> <input class="c" type="radio"> 18</label></p></div><div class="q-explanation"><p>Each and every orbital can hold two electrons. Any s sub-level contains one orbital and can therefore hold two electrons. Any p sub-level contains three orbitals and can therefore hold six electrons. Any d sub-level contains five orbitals and can therefore hold ten electrons. Any f sub-level contains seven orbitals and can therefore hold fourteen electrons.</p><p>The n=3 energy level consists of the 3s, 3p, and 3d sub-levels.</p><p>Thus the answer is <b>18</b> (2+6+10).</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="totals"><span class="score"></span><button class="btn btn-success btn-block text-center check-total"><i class="fa fa-check-square-o"></i> Check</button></div></div><hr> </div> <div class="page-container panel-self-assessment" data-id="2125"> <div class="panel-heading">MY PROGRESS</div> <div class="panel-body understanding-rate"> <div class="msg"></div> <label class="label-lg">Self-assessment</label><p>How much of <strong> Atomic Structure AHL (HL only) paper 1 questions</strong> have you understood?</p><div class="slider-container text-center"><div id="self-assessment-slider" class="sib-slider self-assessment " data-value="1" data-percentage=""></div></div> <label class="label-lg">My notes</label> <textarea name="page-notes" class="form-control" rows="3" placeholder="Write your notes here..."></textarea> </div> <div class="panel-footer text-xs-center"> <span id="last-edited" class="mb-xs-3"> </span> <div class="actions mt-xs-3"> <button id="save-my-progress" type="button" class="btn btn-sm btn-primary text-center btn-xs-block"> <i class="fa fa-fw fa-floppy-o"></i> Save </button> </div> </div></div> <div id="modal-feedback" class="modal fade" tabindex="-1" role="dialog"> <div class="modal-dialog" role="document"> <div class="modal-content"> <div class="modal-header"> <h4 class="modal-title">Feedback</h4> <button type="button" class="close hidden-xs hidden-sm" data-dismiss="modal" aria-label="Close"> <span aria-hidden="true">×</span> </button> </div> <div class="modal-body"> <div class="errors"></div> <p><strong>Which of the following best describes your feedback?</strong></p> <form method="post" style="overflow: hidden"> <div class="form-group"> <div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Recommendation"> Recommend</label></div><div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Problem"> Report a problem</label></div><div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Improvement"> Suggest an improvement</label></div><div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Other"> Other</label></div> </div> <hr> <div class="row"> <div class="col-md-6"> <div class="form-group"> <label for="feedback-name">Name</label> <input type="text" class="form-control" name="feedback-name" placeholder="Name" value=" "> </div> </div> <div class="col-md-6"> <div class="form-group"> <label for="feedback-email">Email address</label> <input type="email" class="form-control" name="feedback-email" placeholder="Email" value="@airmail.cc"> </div> </div> </div> <div class="form-group"> <label for="feedback-comments">Comments</label> <textarea class="form-control" name="feedback-comments" style="resize: vertical;"></textarea> </div> <input type="hidden" name="feedback-ticket" value="082b9c9c4ae3624d"> <input type="hidden" name="feedback-url" value="https://studyib.net/chemistry/page/2125/atomic-structure-ahl-hl-only-paper-1-questions"> <input type="hidden" name="feedback-subject" value="7"> <input type="hidden" name="feedback-subject-name" value="Chemistry"> <div class="pull-left"> </div> </form> </div> <div class="modal-footer"> <button type="button" class="btn btn-primary btn-xs-block feedback-submit mb-xs-3 pull-right"> <i class="fa fa-send"></i> Send </button> <button type="button" class="btn btn-default btn-xs-block m-xs-0 pull-left" data-dismiss="modal"> Close </button> </div> </div> </div></div> <style type="text/css" media="screen">/* Important part */
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