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Sub-Atomic Particles

Boron does no...

Question

Boron does not from $B^{3 +}$ ions whereas $A l$ forms $A l^{3 +} i o n s$ . This because :

A

The size of

B

atom is smaller than that of

A l

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B

The sun of

I E_{1} + I E_{2} + I E_{3}

of

B

is equal to that of

A l

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C

The sum of

I E_{1} + I E_{2} + I E_{3}

of Al is much higher than that of

B

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D

Both A and B

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Solution

The correct option is A The size of $B$ atom is smaller than that of $A l$
B has eletronic configuration ${1 s}^{2} 2 s^{2} 2 p^{1}$
After removal of one electron electronic configuration becomes stable i.e full filled s - orbital.

So second IE becomes very high so, total $I E_{1} + I E_{2} + I E_{3}$ is very high and hence formation of $B^{3 +}$ is impossible

And due to small size of B, it becomes very difficult to withdraw electrons

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Similar questions

{I E}_{1}

{I E}_{2}

{I E}_{3}

{I E}_{4}

N i

P t

are given below

	${I E}_{1} + {I E}_{2}$	${I E}_{3} + {I E}_{4}$
$N i$	$2.49$	$8.80$
$P t$	$2.66$	$6.70$

If true enter 1, else enter 0.
It can be predicted that

{N i}^{2 +}

is more stable than

{P t}^{2 +}

.

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{(I E)}_{2}

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{I E}_{1} {I E}_{2} {I E}_{3} {I E}_{4} {I E}_{5}

of an element are

15.1, 24.3, 34.5, 46.8, 162.2, e V

respectively. This element is likely to be :(I.E.=Ionisation energey )

I E_{2}

for an element is invariably higher than

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{m o I}^{- 1}

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	Z	${(I E)}_{1}$	${(I E)}_{2}$	${(I E)}_{3}$
$L i$	3	520.1	$X (1)$	11810
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From above table

X (1)

Y (2)

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