Question

Lewis's concept of covalency of an element involves octet rule. Later on, it was found that many elements in their compounds, e.g., $Be{F}_2,B{F}_3$ etc. have incomplete octet whereas $PC{l}_5,S{F}_6$, etc. have expanded octet. This classical concept also failed in predicting the geometry of molecules. The modern concept of covalence was proposed in terms of valence bond theory successfully explained the geometry of various molecules but failed in many molecules. The geometry of such molecules was explained by the VSEPR concept. Finally, the molecular orbital theory was proposed to explain many other molecules.

Which are true statements among the following?

(I) $I_3^{+}$ has bent structure
(II) p$\pi -$p$\pi$ bonds are present in $S{O}_2$
(III) $Se{F}_4$ and $C{H}_4$ has same shape
(IV) $Xe{F}_2$ and $C{O}_2$ has same shape

• A. I, II, III, IV
• B. I, II
• C. I, II, IV
• D. None of these

Answer 1

Answer: (C)

I, II, IV

(I) $I_3^{+}$ has a bent structure. The central iodine atom has two lone pairs and two bond pairs. It is $s{p}^3$ hybridized.

(II) $p_{\pi }-p_{\pi }$ bond is present in $S{O}_2$. On p orbital of S atom overlaps with one p orbital of one oxygen atom to form $p_{\pi }-p_{\pi }$ bond. Similarly, another $p_{\pi }-d_{\pi }$ bond is formed with another oxygen atom.

(III) $Se{F}_4$ has 1 lone pair but in $C{H}_4$, there is no lone pair only 4 bond pairs are present. So, the shape of $Se{F}_4$ is see-saw and $C{H}_4$ has a tetrahedral structure.

So, $Se{F}_4$ and $C{H}_4$ has not the same shape.

(IV) $Xe{F}_2$ and $C{O}_2$ has same shape. $Xe{F}_2$ with $s{p}^{3}d$ hybridization and 2 bond pairs and 3 lone pairs have linear geometry. $C{O}_2$ with sp hybridization has 2 double bonds and zero lone pairs and has linear geometry.

So, the correct answer is $C$