Ncert Solutions Chemistry Class 12th
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New answer posted
10 months agoContributor-Level 10
7.10
The balanced equation for the hydrolytic reaction of PCl5 in heavy water (D2O):

Therefore, on adding the above two equation we can say that the net reaction is:

New answer posted
10 months agoContributor-Level 10
7.9
Generally, in CCl4, all the C-Cl bonds are equal. But that is not in the case of PCl5. All the bonds in PCl5 are not similar. It has three equatorial and two axial bonds. Therefore, when PCl5 is heated strongly, the axial bonds breakdown and thus PCl5 decomposes to PCl3.

New question posted
10 months agoNew answer posted
10 months agoContributor-Level 10
7.8
White phosphorous when heated with concentrated NaOH solution (in a CO2 atmosphere) gives phosphine, PH3. Another product is Sodium hypophosphite, NaH2PO2.

New answer posted
10 months agoContributor-Level 10
7.7 Hybridisation of P in PH3 is sp3. Out of the four sp3 hybrid orbitals, three are involved in bonding with the three H-atoms and the fourth one has a lone pair in it.
PH3 when combines with a proton, it forms PH + in which the lone pair is absent. So, here all the four hybrid orbitals are involved in bonding.
Now; the lone pair- bond-pair repulsion is stronger than the corresponding bond pair-bond pair repulsion; thus, the tetrahedral shape associated with sp3 bonding in PH3 is changed to pyramidal and due to the absence of lone pair in PH4+, there is no lone pair-bond pair repulsion. Hence; the bond angle in PH4+ is higher than the b
New answer posted
10 months agoContributor-Level 10
7.6 Diagram of N2O5:

Covalence of nitrogen in N2O5 is 4 (as can be said from the above diagram).
Note: Covalence is the number of electron pairs that an atom can share with other atoms.
New answer posted
10 months agoContributor-Level 10
7.5
Ammonia on reacting with Cu2+ acts as a Lewis base and donates its electron pair to the metal ion and forms a linkage with the metal ion.

Here, colour of Cu2+ is blue whereas the colour of [Cu (NH3)4]2+ is deep blue.
New answer posted
10 months agoContributor-Level 10
7.4
Ammonia (NH3) is generally prepared through a many numbers of processes out of which Haber's process is the most important one. Following the Le Chatlier's principle and taking into considerations the reaction conditions of the Haber's process, the yield of ammonia can be maximized by
High pressure (~ 200 atm)
High temperature (~ 700K)
A mixture of Iron oxide with small amounts of K2O and Al2O3 used as a catalyst. (This mixture acts as a positive catalyst in the Haber's process)
New answer posted
10 months agoContributor-Level 10
7.3
N2 (Dinitrogen) is formed by sharing three electron pairs between two nitrogen atoms, which are joined by a triple bond as shown below:

Due to a complete octet of the nitrogen atoms and bonded by strong triple bonds, due to which its bond dissociation energy is very high. Due to this reason, N2 is very less reactive at room temperature.
New answer posted
10 months agoContributor-Level 10
7.2
Atomic size increases down the group and thus the stability of the hydrides of Group-15 elements decreases. Now, the least stable hydride is the one which is more reactive and possessing higher reducing strength.
The reducing character of hydrides increases on moving from NH3 to BiH3. Hence, BiH3 is the strongest reducing agent among hydrides of Group-15 elements.
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