Gadolinium (Gd) has a $4 f^{7} 5 d^{1} 6 s^{2}$ electronic configuration outside the [Xe] core. Find the spin only magnetic moment of $G d^{3 +}$ .

\sqrt{63} B . M

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\sqrt{35} B . M

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\sqrt{48} B . M

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\sqrt{24} B . M

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Solution

The correct option is A $\sqrt{63} B . M$
$G d^{3 +}$ has an electronic configuration of $[X e] 4 f^{7}$ .

$[X e] 4 f^{7}$ has seven unpaired electrons.

Spin only magnetic moment, $μ = \sqrt{n (n + 2)}$ B.M

$μ = \sqrt{7 (7 + 2)} = \sqrt{63}$ B.M

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Similar questions

Q. Gadolinium (Gd) has a

4 f^{7} 5 d^{1} 6 s^{2}

electronic configuration outside the [Xe] core. Find the spin only magnetic moment of

G d^{3 +}

Q. The lanthanide element that has the electronic configuration

[X e] 4 f^{7} 5 d^{1} 6 s^{2}

is:

The electronic configuration of gadolinium (Atomic number 64) is
(a) $[X e] 4 f^{3} 5 d^{5} 6 s^{2}$
(b) $[X e] 4 f^{7} 5 d^{2} 6 s^{1}$
(c) $[X e] 4 f^{7} 5 d^{1} 6 s^{2}$
(d) $[X e] 4 f^{8} 5 d^{6} 6 s^{2}$

Q. Use Hund's rule to derive the electronic configuration of

C e^{3 +}

ion and calculate its magnetic moment on the basis of spin-only formula.

Q. A compound of vanadium has a spin magnetic moment of 1.73 BM. Work out the electronic configuration of the vanadium ion in the compound.

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Gadolinium (Gd) has a 4f75d16s2 electronic configuration outside the [Xe] core. Find the spin only magnetic moment of Gd3+.

The correct option is A √63B.MGd3+ has an electronic configuration of [Xe]4f7. [Xe]4f7 has seven unpaired electrons. Spin only magnetic moment, μ=√n(n+2) B.M μ=√7(7+2)=√63 B.M

Gadolinium (Gd) has a $4 f^{7} 5 d^{1} 6 s^{2}$ electronic configuration outside the [Xe] core. Find the spin only magnetic moment of $G d^{3 +}$ .

The correct option is A $\sqrt{63} B . M$
$G d^{3 +}$ has an electronic configuration of $[X e] 4 f^{7}$ .

$[X e] 4 f^{7}$ has seven unpaired electrons.

Spin only magnetic moment, $μ = \sqrt{n (n + 2)}$ B.M

$μ = \sqrt{7 (7 + 2)} = \sqrt{63}$ B.M