Chemistry · d and f Block Elements

JEE Main 2024 — 9 April, Shift 2 — Question 74

A transition metal ' M ' among $\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}$ and Fe has the highest second ionisation enthalpy. The spin only

magnetic moment value of $\mathrm{M}^{+}$ ion is $\qquad$ BM (Near integer)

(Given atomic number $\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}$ : 24, $\mathrm{Mn}: 25, \mathrm{Fe}: 26)$

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

Cr has high $I{{E}_{2}}$ among them due to $3{{d}^{5}}$ configuration in crt

B.M $\begin{aligned}& =\sqrt{5\left( 572 \right)} \\ & =\sqrt{35} \\ & \approx 6 \\ \end{aligned}$

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Chemistry
Chapter: d and f Block Elements
Topic: Introduction and Properties of Transition Elements

Answer: 6

Step-by-step solution

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