Question

Following are the configuration of $4$ atoms.
P$=\left[Ne\right]3s^{2}3p^3$
Q$=\left[Ar\right]3d^{10}4s^{2}4p^3$
R$=\left[He\right]2s^{2}2p^5$
S$=\left[Ne\right]3s^1$
Incorrect statement is:

• A. EA : Q $>$ R[EA $=$ electron affinity]
• B. EN : R $>$ P[EN $=$ electronegative]
• C. IP : P $>$ S [IP $=$ Ionisation potential]
• D. Size : S $>$ R

Answer 1

The correct option is A EA : Q $>$ R[EA $=$ electron affinity]

$Q=\left[Ar\right]3d^{10}4s^{2}4p^3$

$R=\left[He\right]2s^{2}2p^5$

$EA:Q>R$ [EA=Electron Affinity]

$Q$ has stable half filled electronic configuration and to add an $e^{-}$ is difficult whereas $EA$ of $R$ will be more than $Q$ as $R$ requires only one more $e^{-}$ to complete its' $p-$ orbital, hence is more liable to add an $e^{-}$ and achieve stable electronic configuration.