Magnetic moment of 25 Mn x+ is √35 B.M. Then the value of x is

Question

Magnetic moment of 25 Mn x+ is √35 B.M. Then the value of x is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

25 Mn -[ Ar ] 3 d 5 4 s 2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/2500350579b23caba587ce8b1cfef7fe-.png" /> Given √n(n+2)=√35 ⇒ n=5 Hence, to have ' 5 ' unpaired electrons Mn must be in ' +2 ' state.

Q. Magnetic moment of $_{25} M n^{x +}$ is $35$ B.M. Then the value of $x$ is

$2 5^{M n} - [A r] 3 d^{5} 4 s^{2}$

Given $n (n + 2) = 35$
$\Rightarrow n = 5$
Hence, to have ' $5$ ' unpaired electrons $M n$ must be in ' $+ 2$ ' state.

Q. Magnetic moment of 25​Mnx+ is 35​ B.M. Then the value of x is

25Mn−[Ar]3d54s2 Given n(n+2)​=35​ ⇒n=5 Hence, to have ' 5 ' unpaired electrons Mn must be in ' +2 ' state.

Q. Magnetic moment of $_{25} M n^{x +}$ is $35$ B.M. Then the value of $x$ is

$2 5^{M n} - [A r] 3 d^{5} 4 s^{2}$

Given $n (n + 2) = 35$
$\Rightarrow n = 5$
Hence, to have ' $5$ ' unpaired electrons $M n$ must be in ' $+ 2$ ' state.