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This is the second of our many quizzes on Bonding. Click on this to access more such competitive quizzes on Physical Chemistry.

Bonding Quiz 2

chemistry

 The molecules that come under the class of isoelectronic species will have the same-

 The number of bonding electrons and antibonding electrons will be the same for the set of molecules that have the same number of electrons.

Isoelectronic species are those species that will have the same number of electrons. And as there is the same number of electrons the number of bonding electrons and antibonding electrons will be the same for the set of isoelectronic species. Hence the bond order will be the same for the set of molecules as the bond order value is half of the value of the difference between bonding and antibonding electrons.

 The hybridization of the second analog member of the alkene family is:

 The second analogue member of the alkene family has a steric number value of 3.

We know that the second analogue of the alkene family is ethane which has the formulae ${{C}_{2}}{{H}_{4}}$. In ethene, the steric number value is 3. Hence the hybridization related to steric number 3 is $s{{p}^{2}}$ hybridization. Three orbitals, one-s orbital and two p orbitals are involved in chemical bonding.

 Identify which type of bonds are present in the following given molecules.<br>a) NaCl   	b) Ethylene   	 c) Benzene	d) Nitrogen 

  The type of bond between the atoms in the molecule is going to depend on the electronegativity difference between the atoms present in the molecule. 

 Among the given compounds only NaCl contains ionic bonds and the remaining compounds contain sigma and pi-bond. Because NaCl only contains the atoms which have an electronegativity difference of more than 2.7 between the atoms. <br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/70ffc80e-5597-4d73-8846-74dc4b470cac2321336892991096570.png"/><br>NaCl- Ionic bond, Ethylene- Sigma and pi-bond, Benzene-Sigma and pi-bond, Nitrogen-Sigma and pi-bond.

 How many lone pairs and bond pairs should be drawn to complete the structure of the $Cl{{F}_{3}}$ molecule?

 In $Cl{{F}_{3}}$, Cl is $s{{p}^{3}}d$ hybridized, and there are seven valence electrons in Cl.

 In the molecule $Cl{{F}_{3}}$, Cl is the central atom and we need three unpaired electrons to bond with three F atoms. The Cl has seven valence electrons and the hybridization of Cl in $Cl{{F}_{3}}$ is $s{{p}^{3}}d$ hybridization. So $Cl{{F}_{3}}$, there will be three bond pairs of electrons and two lone pairs of electrons.

 The given molecular orbital representation is of:<br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/95d1664e-d61c-4fa2-a549-a392317602ee7103660714276725484.png"/>

  The total number of electrons distributed between bonding and antibonding orbitals will be the same as that of the number of electrons present in the carbon molecule.

 The given molecular orbital diagram is of a carbon molecule. The carbon atom has an atomic number of 6 and hence there are 6 electrons in the carbon atom and there will be 12 electrons in the carbon molecule. And the electrons are distributed in bonding and antibonding orbitals.The configuration is:$\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\pi 2{{p}_{x}}^{2}\pi 2{{p}_{y}}^{2}$