The atomic number of vanadium (V), chromium&n | vedclass

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Question

(b)$V = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^3}4{s^2}$

$Cr = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}4{s^1}$

$Mn = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}

Vedclass · Accepted Answer

$Cr$

Vedclass · Answer

(b)$V = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^3}4{s^2}$

$Cr = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}4{s^1}$

$Mn = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}4{s^2}$

$Fe = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^6}4{s^2}$

In second ionization enthalpy $C{r^ + }$ has exact half filled $d-$ sub shell.

The atomic number of vanadium $(V)$, chromium $(Cr)$, manganese $ (Mn)$ and iron $(Fe) $ are respectively $ 23,\, 24, \,25 $ and $ 26 $ which one of these may be expected to have the highest second ionization enthalpy

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