The electronic configuration $[K r] 4 d^{10}$ represents ____
( $Z$ for $S r = 38$ , $S n = 50$ , $T e = 52$ , $A g = 47$ )

S r^{2 +}

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S n^{2 +}

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T e^{2 +}

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A g^{+}

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Solution

The correct option is D $A g^{+}$
(a) $S r^{2 +}$
The atomic number of Strontium is $(S r) = 38$
Electronic configuration of $S r : [K r] 5 s^{2}$
The electronic configuration of $S r^{2 +} (36 electrons) is : [K r]$

(b) $S n^{2 +}$
The atomic number of Tin $(S n) = 50$
Electronic configuration of $S n = [K r] 4 d^{10} 5 s^{2} 5 p^{2}$
The electronic configuration of $S n^{2 +}$ ( with 48 electrons) $= [K r] 4 d^{10} 5 s^{2}$

(c) $T e^{2 +}$
The atomic number of Tellurium $(T e) = 52$
Electronic configuration of $T e = [K r] 4 d^{10} 5 s^{2} 5 p^{4}$
Electronic configuration of $T e^{2 +}$ ( 50 electrons) $= [K r] 4 d^{10} 5 s^{2} 5 p^{2}$

(d) $A g^{+}$
The atomic number of silver $(A g) = 47$
Electronic configuration of $A g = [K r] 4 d^{10} 5 s^{1}$
Electronic configuration of $A g^{+}$ ( with 46 electrons) $= [K r] 4 d^{10}$

Hence, The electronic configuration $[K r] 4 d^{10}$ represents the $A g^{+}$ ion.

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Similar questions

[K r] 4 d^{10}

Z

S r = 38

S n = 50

T e = 52

A g = 47

[K r] 4 d^{10}

A g^{+} (a q .) + e^{-} \to A g (s)

S n^{2 +} (a q .) + 2 e^{-} \to S n (s)

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e m f

S n | S n^{2 +} (1 M) | | A g^{+} (1 M) | A g

Pd has exceptional outer electronic configuration as $4 d^{10} 5 s^{0}$ , it belongs to

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