The formation of PH+4 is difficult compared to NH+4 because
(1) lone pair of phosphorous is optically inert
(2) lone pair of phosphorous resides in almost pure p- orbital
(3) lone pair of phosphorous resides at sp3 orbital
(4) lone pair of phosphorous resides in almost pure s- orbital
(4) +H+ PH+4
According to Drago's rule lone pair on phosphorous resides in almost pure s- orbital, hence due to non directional nature, its overlapping tendency is greatly reduced in comparison to a lone pair present in hybrid orbital, which is directional as present in
It is because inability of ns2 electrons of the valence shell to participate in bonding that
(a) Sn2+ is reducing while Pb4+ is oxidising agent
(b) Sn2+ is oxidising while Pb4+ is reducing agent
(c) Sn2+ and Pb2+ are both oxidising and reducing agent
(d) Sn4+ is reducing while Pb4+ is oxidising agent
(a) The inability of ns2 electrons of the valence shell to participate in bonding is called as inert pair effect. Due to this effect, the lower oxidation state becomes more stable on descending the group. Thus, Sn2+ is a reducing agent while Pb4+ act as an oxidising agent.
Among the following, which one is a wrong statement?
(a) PH5 and BiCl5 do not exist
(b) ρp-dp bonds are present in SO2
(c) SeF4 and CH4 have same shape
(d) has bent geometry
(c) PH5 does not exist due to very less electronegativity difference between P and H. Hydrogen is slightly more electronegative than phosphorus, thus could not hold significantly the sharing electrons.
On the other hand, ВіCl5 does not exist due to inert pair effect. On moving down the group, +5 oxidation state becomes less stable while +3 oxidation state becomes more stable.
In SO2,ρp-dp and pp-pp both types of bonds are present.
Thus, SeF4 and CH4 do not have same shape.
Thus, option (c) is incorrect statement.
When copper is heated with conc. HNO3 it produces
(a) Cu(NO3)2 and NO (b)Cu(NO3)2,NO and NO2
(c)Cu(NO3)2 and N2O (d) Cu(NO3)2 and NO2
(d) Nitric acid acts as an oxidising agent while reacting with copper. When dil. HNO3 reacts, reaction proceeds as:
and when conc. HNO3 is used, reaction proceeds as