Stability decreases in the order

H-F > H-Cl > H-Br > H-I

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H-Cl > H-F > H-Br > H-I

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H-F > H-Br > H-Cl > H-I

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H-F > H-Cl > H-I > H-Br

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Solution

The correct option is A H-F > H-Cl > H-Br > H-I
Because of decrease in bond dissociation enthalpy, the stability of halides decreases down the group. The size of halogen also contributes, thus increase in halogen size decreases stability.
Hence, the answer is H-F > H-Cl > H-Br > H-I

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Similar questions

Q. Number of correct options among the following is/are:

(a) I^{-} > B r^{-} > C l^{-} > F^{-} (Polarizability)

(b) L i^{+} > N a^{+} > K^{+} > R b^{+} (Polarization power)

(c) H_{2} O > H_{2} S > H_{2} S e > H_{2} T e (Boiling point)

(d) H_{2}^{-} > H_{2}^{+} (order of stability)

Which of the following is not correct?

$O r d e r o f b a s i c i t y O r d e r o f n u c l e o p h i l i c i t y (a) F^{-} > C l^{-} > B r^{-} B r_{-} > C l^{-} > F^{-} (b) H - O - O^{-} > O H^{-} H - O - O^{-} > O H^{-} (c) C H_{3} > N H_{2}^{-} > O H - C H_{3} > N H_{2}^{-} > O H^{-} (d) H_{2} O > H_{2} S H_{2} O < H_{2} S$

Given the standard reduction potentials:

Zn²⁺/Zn = - 0.74 V, Cl₂/Cl^- = 1.36 V, H⁺/1/2H₂ = 0 V and Fe³⁺/Fe²⁺ = 0.77 V.
The order of increasing strength as reducing agent is

The correct statement with regard to $H_{2}^{+}$ & $H_{2}^{-}$ is

H_{2}, L i_{2}, B_{2}

each has bond order equal to 1, the order of their stability is :

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Stability decreases in the order

The correct option is A H-F > H-Cl > H-Br > H-I Because of decrease in bond dissociation enthalpy, the stability of halides decreases down the group. The size of halogen also contributes, thus increase in halogen size decreases stability. Hence, the answer is H-F > H-Cl > H-Br > H-I

The correct option is A H-F > H-Cl > H-Br > H-I
Because of decrease in bond dissociation enthalpy, the stability of halides decreases down the group. The size of halogen also contributes, thus increase in halogen size decreases stability.
Hence, the answer is H-F > H-Cl > H-Br > H-I