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Octet Rule

IE1 of O< I...

Question

${I E}_{1}$ of $O < {I E}_{1}$ of $N$ but ${I E}_{2}$ of $O > {I E}_{2}$ of $N$ - Explain

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Solution

$2^{n d}$ I.E of oxygen is more than that of nitrogen because it attains stable half filled $p -$ orbital configuration where as nitrogen is left with only $2$ electrons in its $p - o r b i t a l$ , so it is easier to remove an electron from nitrogen than from oxygen thus oxygen has a greater ionization enthalpy.

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{(I E)}_{1}

{(I E)}_{2}

{m o l}^{- 1}

of K and Ca have been given as

	${(I E)}_{1}$	${(I E)}_{2}$
K	419	3052
Ca	590	1145

The difference can be explained by:

Q. Which represents alkali metals based on

(I E_{1})

(I E_{2})

		$(I E)_{1}$	$(I E)_{2}$
(a)	X	100	110
(b)	Y	95	120
(c)	Z	195	500
(d)	M	200	250

I E_{1}

M g

is more than that of

N a

I E_{2}

value is less. Explain?

{I E}_{2}

for an element are invariably higher than

{I E}_{1}

Q. The ionization enthalpies,

(I E)_{1}

(I E)_{2}

^{- 1}

of a few elements designated by Roman numerals are shown below :

Element	$(I E)_{1}$	$(I E)_{2}$
A	2372	5251
B	520	7300
C	900	1760
D	1680	3380

Which of the above elements form a stable binary halide of the formula

M X_{2}

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