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(A) First,identify the elements:$P$ is Phosphorus $(Z=15)$,$Q$ is Arsenic $(Z=33)$,$R$ is Fluorine $(Z=9)$,and $S$ is Sodium $(Z=11)$.$(A)$ Electron a

Vedclass · Accepted Answer

$EA : Q > R$ (where $EA = 	ext{electron affinity}$)

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(A) First,identify the elements:$P$ is Phosphorus $(Z=15)$,$Q$ is Arsenic $(Z=33)$,$R$ is Fluorine $(Z=9)$,and $S$ is Sodium $(Z=11)$.$(A)$ Electron affinity $(EA)$: For group $15$ elements,$EA$ decreases down the group. Thus,$P > Q$. However,$R$ (Fluorine) has a lower $EA$ than expected due to small size and inter-electronic repulsion,but generally,$EA$ of $P$ and $Q$ are very low compared to halogens. Comparing $Q$ and $R$,$R$ is a halogen and has a higher $EA$ than $Q$. Thus,$Q > R$ is incorrect.$(B)$ Electronegativity $(EN)$: $R$ $(F)$ is the most electronegative element,so $R > P$ is correct.$(C)$ Ionisation potential $(IP)$: $P$ has a stable half-filled $p$-orbital $(3p^3)$,making its $IP$ higher than $S$ $(3s^1)$,which is correct.$(D)$ Size: $S$ $(Na)$ is in period $3$,while $R$ $(F)$ is in period $2$. Thus,$S > R$ is correct.Therefore,the incorrect statement is $A$.

The electronic configurations of $4$ atoms are given below:
$P = [Ne] \, 3s^2 3p^3$
$Q = [Ar] \, 3d^{10} 4s^2 4p^3$
$R = [He] \, 2s^2 2p^5$
$S = [Ne] \, 3s^1$
Which of the following statements is incorrect?

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The electronic configurations of $4$ atoms are given below:$P = [Ne] \, 3s^2 3p^3$$Q = [Ar] \, 3d^{10} 4s^2 4p^3$$R = [He] \, 2s^2 2p^5$$S = [Ne] \, 3s^1$Which of the following statements is incorrect?