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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Chemistry
Chapter
d and f Block Elements

JEE Main 2024 · 9 April · Evening Shift · Chemistry · d and f Block Elements

A transition metal ' ( M ) ' among ( Sc, Ti, V, Cr, Mn ) and ( Fe )...

A transition metal ' (M\mathrm{M}) ' among (Sc,Ti,V,Cr,Mn\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}) and (Fe\mathrm{Fe}) has the highest second ionisation enthalpy. The spin-only magnetic moment value of (\mathrm{M}^{+})ion is _______. (BM\mathrm{BM}) (Near integer) (Given atomic number (Sc\mathrm{Sc}: 21, Ti\mathrm{Ti}: 22,  V\mathrm{~V}: 23, Cr\mathrm{Cr}: 24, Mn\mathrm{Mn}: 25, Fe\mathrm{Fe}: 26))

Correct answer

0

Step-by-step solution

Among given metals, Cr\mathrm{Cr} has maximum IE2\mathrm{IE}_2 because Second electron is removed from stable configuration 3d5Cr+:[Ar]3 d54 s0∴3 d^5\mathrm{Cr}^{+}:[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0\therefore No of unpaired e−\mathrm{e}^{-}in Cr+\mathrm{Cr}^{+}is 5,n=55, \mathrm{n}=5 So, Magnetic moment =n(n+2) B.M =5(5+2)=5.92BM≈6\begin{aligned} \text{ So, Magnetic moment } & =\sqrt{\mathrm{n}(\mathrm{n}+2)} \text{ B.M } \\ & =\sqrt{5(5+2)}=5.92 \quad \mathrm{BM} \approx 6 \end{aligned}$$
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