NEET Exam  >  NEET Notes  >  Chemistry Class 12  >  DPP for NEET: Daily Practice Problems, Ch: P-block Elements (Group 15,16,17 and 18) (Solutions)

P-block Elements (Group 15,16,17 and 18) Practice Questions - DPP for NEET

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 Page 1


1. (a) Hypophosphorous acid is H
3
PO
2
 in which O.S. of P is +1
2. (b) The bond angle decreases on moving down the group due to
decrease in bond pair-bond pair repulsion.
NH
3
  PH
3
 AsH
3
SbH
3
 BiH
3
107º 94º   92º   91º  90º
This can also be explained by the fact that as the size of central atom
increases sp
3
 hybrid orbital becomes more distinct with increasing
size of central atom i.e. pure p-orbitals are utilized in M–H
bonding.
3. (b)  
       Blue
4. (c) The nitroprusside ion is [Fe(CN)
6
NO
+
]
2–
. The magnetic moment
measurements reveal the presence of 4 unpaired electrons in Fe
which must be then in Fe
++
 (3d
6
) and not Fe
+++
(3d
5
)
5. (c) [Fe(H
2
O)
5
NO]
2+
 ion is formed
6. (a) 4HCl + O
2
 2Cl
2
 + 2H
2
O
           air   cloud of white fumes
7. (a) ICl
7
. The hybridisation is 
8. (a) Due to large enthalpy of vaporisation SO
2
 can be used as
refrigerant
9. (c) Oxides which are more ionic in nature (salt - like)  are known as
saline oxides e.g. oxides of alkali metals
10. (b) Among  the given compounds, the is most basic. Hence has
highest proton affinity
11. (c) Helium is heavier than hydrogen although it is non-inflammable
12. (b) it can form two series of salts by replacement of H
Page 2


1. (a) Hypophosphorous acid is H
3
PO
2
 in which O.S. of P is +1
2. (b) The bond angle decreases on moving down the group due to
decrease in bond pair-bond pair repulsion.
NH
3
  PH
3
 AsH
3
SbH
3
 BiH
3
107º 94º   92º   91º  90º
This can also be explained by the fact that as the size of central atom
increases sp
3
 hybrid orbital becomes more distinct with increasing
size of central atom i.e. pure p-orbitals are utilized in M–H
bonding.
3. (b)  
       Blue
4. (c) The nitroprusside ion is [Fe(CN)
6
NO
+
]
2–
. The magnetic moment
measurements reveal the presence of 4 unpaired electrons in Fe
which must be then in Fe
++
 (3d
6
) and not Fe
+++
(3d
5
)
5. (c) [Fe(H
2
O)
5
NO]
2+
 ion is formed
6. (a) 4HCl + O
2
 2Cl
2
 + 2H
2
O
           air   cloud of white fumes
7. (a) ICl
7
. The hybridisation is 
8. (a) Due to large enthalpy of vaporisation SO
2
 can be used as
refrigerant
9. (c) Oxides which are more ionic in nature (salt - like)  are known as
saline oxides e.g. oxides of alkali metals
10. (b) Among  the given compounds, the is most basic. Hence has
highest proton affinity
11. (c) Helium is heavier than hydrogen although it is non-inflammable
12. (b) it can form two series of salts by replacement of H
attached to oxygen
13. (d) In gaseous state the HCl is covalent in nature while in aqueous
solution it ionises to give H
+
 and  ions
14. (c) Hypophosphorous acid 
Two H-atoms are attached to P atom.
15. (b) During disproportionation same compound undergo simultaneous
oxidation and reduction.
 redution
16. (b) XeOF
4
 is square pyramidal.
17. (c) Yellow ammonium sulphide is 
18. (c) The correct order of increasing bond angle is
          
In  there are 2 lone pairs of electrons present on the central
chlorine atom. Therefore the bond angle in  is less than 118°
which is the bond angle in ClO
2
 which has less number of
electrons on central chlorine atom.
19. (d) The products of the concerned reaction react each other forming
back the reactants.
.
20. (b) The hybridization of XeO
3
F
2
 is sp
3
d and its structure is trigonal
bipyramidal in which oxygen atoms are situated on the plane and
the fluoride atoms are on the top and bottom.
Page 3


1. (a) Hypophosphorous acid is H
3
PO
2
 in which O.S. of P is +1
2. (b) The bond angle decreases on moving down the group due to
decrease in bond pair-bond pair repulsion.
NH
3
  PH
3
 AsH
3
SbH
3
 BiH
3
107º 94º   92º   91º  90º
This can also be explained by the fact that as the size of central atom
increases sp
3
 hybrid orbital becomes more distinct with increasing
size of central atom i.e. pure p-orbitals are utilized in M–H
bonding.
3. (b)  
       Blue
4. (c) The nitroprusside ion is [Fe(CN)
6
NO
+
]
2–
. The magnetic moment
measurements reveal the presence of 4 unpaired electrons in Fe
which must be then in Fe
++
 (3d
6
) and not Fe
+++
(3d
5
)
5. (c) [Fe(H
2
O)
5
NO]
2+
 ion is formed
6. (a) 4HCl + O
2
 2Cl
2
 + 2H
2
O
           air   cloud of white fumes
7. (a) ICl
7
. The hybridisation is 
8. (a) Due to large enthalpy of vaporisation SO
2
 can be used as
refrigerant
9. (c) Oxides which are more ionic in nature (salt - like)  are known as
saline oxides e.g. oxides of alkali metals
10. (b) Among  the given compounds, the is most basic. Hence has
highest proton affinity
11. (c) Helium is heavier than hydrogen although it is non-inflammable
12. (b) it can form two series of salts by replacement of H
attached to oxygen
13. (d) In gaseous state the HCl is covalent in nature while in aqueous
solution it ionises to give H
+
 and  ions
14. (c) Hypophosphorous acid 
Two H-atoms are attached to P atom.
15. (b) During disproportionation same compound undergo simultaneous
oxidation and reduction.
 redution
16. (b) XeOF
4
 is square pyramidal.
17. (c) Yellow ammonium sulphide is 
18. (c) The correct order of increasing bond angle is
          
In  there are 2 lone pairs of electrons present on the central
chlorine atom. Therefore the bond angle in  is less than 118°
which is the bond angle in ClO
2
 which has less number of
electrons on central chlorine atom.
19. (d) The products of the concerned reaction react each other forming
back the reactants.
.
20. (b) The hybridization of XeO
3
F
2
 is sp
3
d and its structure is trigonal
bipyramidal in which oxygen atoms are situated on the plane and
the fluoride atoms are on the top and bottom.
21. (a)
22. (c) 
 
23. (d)
In case of oxyacids of similar element as the oxidation number of the
central atom increases, strength of acid also increases.
24. (b) The molecular geometry of BrF
5
 is square pyramidal with
asymmetric charge distribution on the central atom.
25. (d)
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