In the hydrides of the nitrogen group,the $H-M-H$ bond angle gradually approaches $90^o$ as we move from $N$ to $Sb$. This indicates that gradually.....
- AThe basic character of the hydrides increases.
- BAlmost pure $p$-orbitals are used in $M-H$ bonds.
- CThe bond strength of the $M-H$ bond increases.
- DThe electron bond pair moves closer to the central atom.
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Similar Questions
Enlist the main uses of nitrogen.
Medium
View Solution$HNO_3$ is produced in good yield by which of the following methods?
Easy
View SolutionThe correct code for stability of oxidation states for given cations is:
$(i)$ $Pb^{2+} > Pb^{4+}, Tl^{+} < Tl^{3+}$
$(ii)$ $Bi^{3+} < Sb^{3+}, Sn^{2+} < Sn^{4+}$
$(iii)$ $Pb^{2+} > Pb^{4+}, Bi^{3+} > Bi^{5+}$
$(iv)$ $Tl^{3+} < In^{3+}, Sn^{2+} > Sn^{4+}$
$(v)$ $Sn^{2+} < Pb^{2+}, Sn^{4+} > Pb^{4+}$
$(vi)$ $Sn^{2+} < Pb^{2+}, Sn^{4+} < Pb^{4+}$
$(i)$ $Pb^{2+} > Pb^{4+}, Tl^{+} < Tl^{3+}$
$(ii)$ $Bi^{3+} < Sb^{3+}, Sn^{2+} < Sn^{4+}$
$(iii)$ $Pb^{2+} > Pb^{4+}, Bi^{3+} > Bi^{5+}$
$(iv)$ $Tl^{3+} < In^{3+}, Sn^{2+} > Sn^{4+}$
$(v)$ $Sn^{2+} < Pb^{2+}, Sn^{4+} > Pb^{4+}$
$(vi)$ $Sn^{2+} < Pb^{2+}, Sn^{4+} < Pb^{4+}$
Nitrous oxide is known as
Easy
View SolutionThe compound$(s)$ which generate$(s)$ $N_2$ gas upon thermal decomposition below $300^{\circ} C$ is (are)
$(A)$ $NH_4 NO_3$
$(B)$ $(NH_4)_2 Cr_2 O_7$
$(C)$ $Ba(N_3)_2$
$(D)$ $Mg_3 N_2$
$(A)$ $NH_4 NO_3$
$(B)$ $(NH_4)_2 Cr_2 O_7$
$(C)$ $Ba(N_3)_2$
$(D)$ $Mg_3 N_2$
MediumIIT 2018
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