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#555" title="Suggested study time: 30 minutes"><i class="fa fa-clock-o"></i> 30'</span> </ol> <article id="main-article"> <p><strong>Topics 14.1 and 14.2</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="7-811-2703"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which contain delocalised electrons?</p><p><strong>1: </strong>C<sub>6</sub>H<sub>5</sub>CH<sub>3</sub></p><p><strong>2: </strong>CO<sub>3</sub><sup>2−</sup></p><p><strong>3:</strong> HCOO<sup>−</sup></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> 1 and 3 only</label></p><p><label class="radio"> <input type="radio"> 2 and 3 only<span></span></label></p><p><label class="radio"> <input type="radio"> 1 and 2 only</label></p><p><label class="radio"> <input class="c" type="radio"> 1, 2 and 3</label></p></div><div class="q-explanation"><p>Species exhibit delocalised electrons (or resonance) when they cannot be represented by a single Lewis diagram and double bonds can be pushed around the structures (see example below).</p><p>The benzene ring (phenyl group C<sub>6</sub>H<sub>5</sub>) in methyl benzene, C<sub>6</sub>H<sub>5</sub>CH<sub>3</sub>, does have delocalised electrons (and exhibits resonance). The carbonate ion and the methanoate ion also both have delocalised electrons (and exhibit resonance). </p><p><b>1, 2 and 3</b> is therefore the correct answer.</p><p>E.g. delocalisation/resonance in carbonate ion:</p><p><img alt="" src="../../images/chemical-bonding-and-structure/carbonate-resonance.png" style="width: 360px; height: 77px;"></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 correctly describes the molecular geometry in BrF<sub>3</sub>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Trigonal planar</span></label> </p><p><label class="radio"> <input type="radio"> <span>Trigonal bipyramidal</span></label></p><p><label class="radio"> <input type="radio"> <span>Trigonal pyramidal</span></label></p><p><label class="radio"> <input class="c" type="radio"> <span>T–shaped</span></label></p></div><div class="q-explanation"><p>Bromine trifluoride, with three bonding pairs and two lone-pairs around the central bromine atom is <strong>T–shaped</strong>, which is therefore the correct answer.</p><p>The shape of the <strong>electron domains</strong> is trigonal bi-pyramidal, but with two equatorial (rather than axial) vertices missing (the lone pairs) the bonds form the shape of a 'T'.</p><p><img alt="" src="../../images/chemical-bonding-and-structure/screenshot-2020-07-23-at-13.47.30.png" style="width: 160px; height: 186px;"></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 correctly describes the <strong>electron domain geometry</strong> and the <strong>molecular geometry</strong>, respectively, in BrCl<sub>4</sub><sup>− </sup>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Octahedral and Tetrahedral</span></label></p><p><label class="radio"> <input class="c" type="radio"> Octahedral and <span>Square planar</span></label></p><p><label class="radio"> <input type="radio"> <span>Tetrahedral and Square planar</span></label></p><p><label class="radio"> <input type="radio"> <span>Tetrahedral and Tetrahedral</span></label></p></div><div class="q-explanation"><p>BrCl<sub>4</sub><sup>−</sup> with four bonding pairs and two lone-pairs around the central bromine atom is <strong>square planar </strong>in molecular shape.</p><p>The shape of the <strong>electron domains</strong> is <strong>octahedral</strong>, but with two vertices missing (the lone pairs) the bonds form the shape of a square.</p><p>Thus, <strong>Octahedral and Square planar </strong>is the correct answer.</p><p><img alt="" src="../../images/chemical-bonding-and-structure/screenshot-2020-07-23-at-16.46.43.png" style="width: 160px; height: 79px;"></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 atom is sp<sup>2</sup> hybridized?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> Nitrogen in NH<sub>3 </sub></label></p><p><label class="radio"> <input type="radio"> Carbon in CO<sub>2</sub></label></p><p><label class="radio"> <input class="c" type="radio"> Carbon in H<sub>2</sub>CO </label></p><p><label class="radio"> <input type="radio"> Oxygen in CO</label></p></div><div class="q-explanation"><p>When identifying the hybridization of a given atom, it is best to use our 'rule-of-thumb'; that the geometry of the<strong> electron domains</strong> (see molecular shapes in the covalent structure section 4.3) tells us the hybridization:</p><p><strong>Tetrahedral electron domains means sp<sup>3</sup> hybridization.</strong></p><p><strong>Trigonal planer electron domains means sp<sup>2</sup> hybridization.</strong></p><p><strong>Linear electron domains means sp hybridization.</strong></p><p>The carbon atom in H<sub>2</sub>CO has electron domains that are arranged in a trigonal planar manner (two single bonds and a double bond) so it must be sp<sup>2</sup> hybridized.</p><p><strong>Carbon in H<sub>2</sub>CO</strong> is therefore the correct answer.</p><p>(In CO<sub>2</sub> the electron domain shape around carbon is linear, so sp hybridized; In CO the electron domain shape around oxygen is linear - a triple bond - so sp hybridized; In NH<sub>3</sub> the electron domain shape around nitrogen is tetrahedral, so sp<sup>3</sup> hybridized.)</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>How many carbon atoms (grey) are sp<sup>3</sup>, sp<sup>2</sup> and sp hybridized in the molecule shown?</p><p><img alt="" src="../../images/organic-chemistry/cyanopropanoate.png" style="width: 240px; height: 110px;"></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> 2 sp<sup>3</sup> carbons; 2 sp<sup>2</sup> carbons; 0 sp carbons</label></p><p><label class="radio"> <input type="radio"> 1 sp<sup>3</sup> carbon; 2 sp<sup>2</sup> carbons; 1 sp carbon</label></p><p><label class="radio"> <input type="radio"> 1 sp<sup>3</sup> carbon; 1 sp<sup>2</sup> carbon; 2 sp carbons</label></p><p><label class="radio"> <input class="c" type="radio"> 2 sp<sup>3</sup> carbons; 1 sp<sup>2</sup> carbon; 1 sp carbon</label></p></div><div class="q-explanation"><p>When identifying the hybridization of a given atom, it is best to use our 'rule-of-thumb'; that the geometry of the<strong> electron domains</strong> (see molecular shapes in the covalent structure section 4.3) tells us the hybridization:</p><p><strong>Tetrahedral electron domains means sp<sup>3</sup> hybridization.</strong></p><p><strong>Trigonal planar electron domains means sp<sup>2</sup> hybridization.</strong></p><p><strong>Linear electron domains means sp hybridization.</strong></p><p>In order as shown, the electron domain shapes around the carbon atoms are: tetrahedral; tetrahedral; trigonal planar; linear. Thus, the hybridization is sp<sup>3</sup>; sp<sup>3</sup>; sp<sup>2</sup>; sp.</p><p><strong>2 sp<sup>3</sup> carbons; 1 sp<sup>2</sup> carbon; 1 sp carbon</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>How many sigma (σ) and pi (π) bonds are there in ethanitrile (CH<sub>3</sub>CN)?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> 4<span> sigma and 3 pi</span></label> </p><p><label class="radio"> <input type="radio"> <span>6 sigma and 1 pi</span></label> </p><p><label class="radio"> <input type="radio"> 4<span> sigma and 2 pi</span></label> </p><p><label class="radio"> <input class="c" type="radio"> 5<span> sigma and 2 pi </span></label> </p></div><div class="q-explanation"><p><img alt="" src="../../images/organic-chemistry/ethanitrile.png" style="width: 240px; height: 114px;"></p><p>There are 7 bonds in the molecule. Every bond is one pair of electrons. Every sigma bond and every pi bond is one pair of electrons.</p><p>A single bond is just a sigma bond.</p><p>A double bond is one sigma bond and one pi bond.</p><p>A triple bond is one sigma and two pi bonds.</p><p>The correct answer is therefore<strong> 5 sigma bonds and 2 pi bonds </strong>present in ethanitrile.</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>What is the hybridization of the oxygen and two nitrogen atoms circled?</p><p><img alt="" src="../../images/organic-chemistry/labelled-hybridization.png" style="width: 360px; height: 211px;"></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> 1st N is sp<sup>2</sup> ; 2nd N is sp<sup>2</sup> ; O is sp<sup>2</sup></label></p><p><label class="radio"> <input type="radio"> 1st N is sp<sup>3</sup> ; 2nd N is sp<sup>3</sup> ; O is sp<sup>2</sup></label></p><p><label class="radio"> <input class="c" type="radio"> 1st N is sp<sup>2</sup> ; 2nd N is sp<sup>3</sup> ; O is sp<sup>3</sup></label></p><p><label class="radio"> <input type="radio"> 1st N is sp ; 2nd N is sp<sup>2</sup> ; O is sp</label></p></div><div class="q-explanation"><p>When identifying the hybridization of a given atom, it is best to use our 'rule-of-thumb'; that the geometry of the<strong> electron domains</strong> (see molecular shapes in the covalent structure section 4.3) tells us the hybridization:</p><p><strong>Tetrahedral electron domains means sp<sup>3</sup> hybridization.</strong></p><p><strong>Trigonal planar electron domains means sp<sup>2</sup> hybridization.</strong></p><p><strong>Linear electron domains means sp hybridization.</strong></p><p>In order as shown:</p><p>1st N electron domain shape is trigonal planar (don't forget the N atom has a lone pair). Thus, the hybridization is sp<sup>2</sup>.</p><p>2nd N electron domain shape is tetrahedral (don't forget the N atom has a lone pair). Thus, the hybridization is sp<sup>3</sup>.</p><p>O electron domain shape is tetrahedral (don't forget the O atom has two lone pairs). Thus, the hybridization is sp<sup>3</sup>.</p><p><strong>1st N is sp<sup>2</sup> ; 2nd N is sp<sup>3</sup> ; O is sp<sup>3</sup> </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>Which species has a square pyramidal molecular geometry?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>BrF<sub>4</sub><sup>−</sup></span></label></p><p><label class="radio"> <input type="radio"> PF<sub>5</sub></label></p><p><label class="radio"> <input type="radio"> SiH<sub>4</sub></label></p><p><label class="radio"> <input class="c" type="radio"> BrF<sub>5</sub></label></p></div><div class="q-explanation"><p>Sketching the Lewis structures will allow us to determine that:</p><p>Bromine pentafluoride has a square pyramidal molecular geometry and <strong>BrF<sub>5</sub></strong> is therefore the correct answer.</p><p><img alt="" src="../../images/chemical-bonding-and-structure/screenshot-2020-07-23-at-16.44.02.png" style="width: 160px; height: 124px;"></p><p>BrF<sub>4</sub><sup>−</sup> is square planar.</p><p>PF<sub>5</sub> is trigonal bipyramidal.</p><p>SiH<sub>4</sub> is tetrahedral.</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>How many sigma (σ) and pi (π) bonds are present in this molecule?</p><p><img alt="" src="../../images/chemical-bonding-and-structure/cresol.png" style="width: 120px; height: 121px;"></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> 12<span> sigma and 3 pi </span></label> </p><p><label class="radio"> <input type="radio"> 11<span> sigma and 4 pi</span></label></p><p><label class="radio"> <input type="radio"> 9<span> sigma and 3 pi</span></label> </p><p><label class="radio"> <input type="radio"> 8<span> sigma and 4 pi</span></label> </p></div><div class="q-explanation"><p>Don't forget that there are single bonds between the OH and the CH<sub>3</sub> atoms.</p><p>There are 9 single bonds and 3 double bonds.</p><p>So, there are 15 bonds in the molecule; every bond is one pair of electrons. Every sigma bond and every pi bond is one pair of electrons.</p><p>A single bond is just a sigma bond.</p><p>A double bond is one sigma bond and one pi bond.</p><p>A triple bond is one sigma and two pi bonds.</p><p>The correct answer is therefore<strong> 12 sigma bonds and 3 pi bonds </strong>present in the molecule.</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>Which species does not have a bond angle of 90°?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> GeF<sub>4</sub></label></p><p><label class="radio"> <input type="radio"> SF<span style="font-size: 11.25px;">6</span></label></p><p><label class="radio"> <input type="radio"> Xe<span>F<sub>4</sub></span></label></p><p><label class="radio"> <input type="radio"> SF<span style="font-size: 11.25px;">4</span></label></p></div><div class="q-explanation"><p>Sketching the Lewis structures will allow us to determine that:</p><p>SF<sub>6</sub> is octahedral; SF<sub>4</sub> has a seesaw shape; XeF<sub>4</sub> is square planar. These all have (approx.) 90° bond angles.</p><p>GeF<sub>4</sub> is tetrahedral (since Ge is in group 14) and so will have 109.5° bond angles, and not have a bond angle of 90°.</p><p><strong>GeF<sub>4</sub></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="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="2703"> <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> Bonding and 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/2703/bonding-and-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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