NCERT Class XI Chemistry Hydrogen Solutions
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Question : 16
Total: 36
Arrange the following :
(i)C a H 2 , B e H 2 T i H 2
(ii) LiH, NaH and CsH in order of increasing ionic character.
(iii) H – H, D – D and F – F in order of increasing bond dissociation enthalpy.
(iv) NaH,M g H 2 H 2 O
(i)
(ii) LiH, NaH and CsH in order of increasing ionic character.
(iii) H – H, D – D and F – F in order of increasing bond dissociation enthalpy.
(iv) NaH,
Solution:
(i) B e H 2 C a H 2 T i H 2
B e H 2 C a H 2 T i H 2
(ii) Electronegativity decreases as Li > Na > Cs. Thus, increasing ionic character : LiH < NaH < CsH.
(iii) Due to lone pairs of F, bond pairs experience repulsion, hence, F–F has low bond dissociation energy. In D–D, due to higher nuclear attraction bond dissociation energy is geater than H– H. Increasing bond dissociation enthalpy : F–F < H–H < D–D.
(iv) NaH is ionic hydride.M g H 2 H 2 O H 2 O
H2O <M g H 2
(ii) Electronegativity decreases as Li > Na > Cs. Thus, increasing ionic character : LiH < NaH < CsH.
(iii) Due to lone pairs of F, bond pairs experience repulsion, hence, F–F has low bond dissociation energy. In D–D, due to higher nuclear attraction bond dissociation energy is geater than H– H. Increasing bond dissociation enthalpy : F–F < H–H < D–D.
(iv) NaH is ionic hydride.
H2O <
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