Search any question & find its solution
Question:
Answered & Verified by Expert
A transition metal ' \(\mathrm{M}\) ' among \(\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}\) and \(\mathrm{Fe}\) has the highest second ionisation enthalpy. The spin-only magnetic moment value of \(\mathrm{M}^{+}\)ion is _______. \(\mathrm{BM}\) (Near integer)
(Given atomic number \(\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26\))
(Given atomic number \(\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26\))
Solution:
2759 Upvotes
Verified Answer
The correct answer is:
6
Among given metals, $\mathrm{Cr}$ has maximum $\mathrm{IE}_2$
because Second electron is removed from stable configuration $3 d^5$
$\mathrm{Cr}^{+}:[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0$
$\therefore$ No of unpaired $\mathrm{e}^{-}$in $\mathrm{Cr}^{+}$is $5, \mathrm{n}=5$
$\begin{aligned}
\text { So, Magnetic moment } & =\sqrt{\mathrm{n}(\mathrm{n}+2)} \text { B.M } \\
& =\sqrt{5(5+2)}=5.92 \quad \mathrm{BM} \approx 6
\end{aligned}$
because Second electron is removed from stable configuration $3 d^5$
$\mathrm{Cr}^{+}:[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0$
$\therefore$ No of unpaired $\mathrm{e}^{-}$in $\mathrm{Cr}^{+}$is $5, \mathrm{n}=5$
$\begin{aligned}
\text { So, Magnetic moment } & =\sqrt{\mathrm{n}(\mathrm{n}+2)} \text { B.M } \\
& =\sqrt{5(5+2)}=5.92 \quad \mathrm{BM} \approx 6
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.