Question

Which of the following does not obey the octet or duplet rule?
$\left(i\right)H_{2}O\left(ii\right)B{F}_3\left(iii\right)PC{l}_5\left(iv\right)C{O}_2\left(v\right)N{H}_3$

• A. $\left(i\right),\left(ii\right),\left(iii\right)$
• B. $\left(ii\right),\left(iii\right)$
• C. $\left(ii\right),\left(iii\right),\left(iv\right)$
• D. $\left(i\right),\left(iv\right),\left(iii\right)$

Answer 1

The correct option is B $\left(ii\right),\left(iii\right)$

(i) In $H_{2}O$, $O$ atom has 8 electrons in its valence shell and $H$ atoms have two electrons.

(ii) The electronic configuration of $B$ is:
$1s^{2}2s^{2}2p^1$. In $B{F}_3$, each $F$ atom shares one electron with boron atom. Thus, boron atom in $B{F}_3$ has 6 valence electrons. So, it does not satisfy the octet rule.

(iii) The electronic configuration of $P$ is: $1s^{2}2s^{2}2p^{6}3s^{2}3p^3$. In $PC{l}_5$, each $Cl$ atom shares one electron with $P$ - atom. Thus, phosphorous atom in $PC{l}_5$ has 10 valence electrons, so it does not satisfy the octet rule.


(iv) In $C{O}_2$, both $C$ and $O$ have 8 valence electrons. Thus, it satisfies octet rule.


(v) In $N{H}_3$, $N$ -atom has 8 valence electrons and $H$ has two valence electrons.

Thus, $B{F}_3$ and $PC{l}_5$ do not follow octet rule.