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Standard XII

Chemistry

Trend in Ionisation Enthalpy

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Question

Find the formula of halide of a metal whose successive ionization enthalpies are x, 2x, 5x, 100x kJ ${m o l}^{- 1}$ respectively :

M X

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{M X}_{2}

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{M X}_{3}

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M_{2} X

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Solution

The correct option is C ${M X}_{3}$
The successive ionization energy is increasing for metal but there is a sudden increase in fourth ionization energy as compared to the regular increasing trend.

This shows that metal is extra stable in +3 oxidation state due to which fourth ionization energy is abnormally high.

As we know that halogens have valency -1 when they react with metals with lesser electronegativity than that of halogens, the metal is most likely to form $M X_{3}$ , since +3 oxidation state of metal is most stable.

Therefore, the correct option is C.

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Similar questions

Q. First and second ionization enthalpies (in kJ/mol) of few elements are given below:

Element	$E I_{I}$	$E I_{I I}$
(i)	520	980
(ii)	900	1760
(iii)	1680	3380
(iv)	2080	3963

Which of the above elements will form halides with formula

M X_{2}

Q. The ionization enthalpies,

(I E)_{1}

and the

(I E)_{2}

in kJ mol

^{- 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}

Q. The first

(Δ_{i} H_{1})

and the second

(Δ_{i} H_{2})

ionization enthalpies (in kJ mol

^{- 1}

) and the

(Δ_{e g} H)

electron gain enthalpy (in kJ mol

^{- 1}

) of a few elements are given below:

Elements	$Δ H_{1}$	$Δ H_{2}$	$Δ_{e g}$ H
I	520	7300	-60
II	419	3051	-48
III	1681	3374	-328
IV	1008	1846	-295
V	2372	5251	+48
VI	738	1451	-40

Which of the above elements is likely to be :
(a) the least reactive element,
(b) the most reactive metal,
(c) the most reactive non-metal,
(d) the least reactive non-metal,
(e) the metal which can form a stable binary halide of the formula MX

_{2}

(X=halogen) and
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?

Q. What would be the formula of metal halide if successive ionisation energies of metal are 5eV, 50eV, 68eV, 90 eV and 120eV?

The first (Δ_iH1) and the second (Δ_iH) ionization enthalpies (in kJ mol^–1) and the (Δ_egH) electron gain enthalpy (in kJ mol^–1) of a few elements are given below:

Elements	Δ_iH	Δ_iH	Δ_egH
I	520	7300	–60
II	419	3051	–48
III	1681	3374	–328
IV	1008	1846	–295
V	2372	5251	+48
VI	738	1451	–40

Which of the above elements is likely to be :

(a) the least reactive element.

(b) the most reactive metal.

(d) the least reactive non-metal.

(e) the metal which can form a stable binary halide of the formula MX₂, (X=halogen).

(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?

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Find the formula of halide of a metal whose successive ionization enthalpies are x, 2x, 5x, 100x kJ mol−1 respectively :

Find the formula of halide of a metal whose successive ionization enthalpies are x, 2x, 5x, 100x kJ ${m o l}^{- 1}$ respectively :