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प्रश्न
The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of sp3 hybridisation in NH3 and only s−p bonding between hydrogen and other elements of the group].
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उत्तर १
Hydride NH3 PH3 AsH3 SbH3
H−M−H angle 107° 92° 91° 90°
The above trend in the H−M−H bond angle can be explained on the basis of the electronegativity of the central atom. Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle. We know that electronegativity decreases on moving down a group. Consequently, the repulsive interactions between the electron pairs decrease, thereby decreasing the H−M−H bond angle.
उत्तर २
In all these cases, the central atom is sp3 hybridized. Three of the four sp3 orbitals form three σ-bonds, while the fourth contains the lone pair of electrons. On moving down from N to Sb, the electronegativity of the central atom goes on decreasing. As a result of this, bond pairs of electrons lie away and away from the central atom. This is because of the force of repulsion between the adjacent bond pairs goes on decreasing and the bond angles keep on decreasing from NH3to SbH3. Thus, bond angles are in the order:
HNH > HPH > HasH > HSbH
(107.8º) (93.6º) (91.8º) (91.3º)
संबंधित प्रश्न
Account for the following: Oxygen shows catenation behavior less than sulphur.
a. Explain the trends in the following properties with reference to group 16:
1 Atomic radii and ionic radii
2 Density
3 ionisation enthalpy
4 Electronegativity
b. In the electolysis of AgNO3 solution 0.7g of Ag is deposited after a certain period of time. Calulate the quantity of electricity required in coulomb. (Molar mass of Ag is 107.9g mol-1)
Which of the following does not react with oxygen directly?
Zn, Ti, Pt, Fe
Why does NH3 form hydrogen bond but PH3 does not?
Knowing the electron gain enthalpy values for \[\ce{O -> O-}\] and \[\ce{O -> O^{2-}}\] as −141 and 702 kJ mol−1 respectively, how can you account for the formation of a large number of oxides having O2− species and not O−?
(Hint: Consider lattice energy factor in the formation of compounds).
Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Give reasons Thermal stability decreases from H2O to H2Te.
Arrange the following in the order of the property indicated against set :
H2O, H2S, H2Se, H2Te − increasing acidic character.
Explain the following properties of group 16 elements :
1) Electro negativity
2) Melting and boiling points
3) Metallic character
4) Allotropy
Give reactions for the following:
O – O single bond is weaker than S – S single bond.
The formation of \[\ce{O^+_2[PtF6]^-}\] is the basis for the formation of first xenon compound. This is because ____________.
Which of the following statement is incorrect?
Given below are two statements labelled as Assertion (A) and Reason (R).
Assertion (A): Electron gain enthalpy of oxygen is less than that of Flourine but greater than Nitrogen.
Reason (R): Ionisation enthalpies of the elements follow the order Nitrogen > Oxygen > Fluorine.
Select the most appropriate answer from the options given below:
Which of the following statements are correct?
(i) \[\ce{CaF2 + H2SO4 -> CaSO4 + 2HF}\]
(ii) \[\ce{2HI + H2SO4 -> I2 + SO2 + 2H2O}\]
(iii) \[\ce{Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O}\]
(iv) \[\ce{Nacl + H2SO4 -> NaHSO4 + HCl}\]
Out of \[\ce{H2O}\] and \[\ce{H2S}\], which one has higher bond angle and why?
In forming (i) \[\ce{N2 -> N^{+}2}\] and (ii) \[\ce{O2 -> O^{+}2}\]; the electrons respectively are removed from:
______ is a gaseous element of group 16.
Given below are two statements:
Statement I: The boiling point of hydrides of Group 16 elements follows the order:
H2O > H2Te > H2Se > H2S
Statement II: On the basis of molecular mass, H2O is expected to have a lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H2O, it has a higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
