NEET Test Series from KOTA - 10 Papers In MS WORD
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11. THE P-BLOCK ELEMENTS
189635
Aluminium chloride exists as dimer, \(A{l_2}C{l_6}\) in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives
1 \({[Al{(OH)_6}]^{3 - }} + 3HCl\)
2 \({[Al{({H_2}O)_6}]^{3 + }} + 3HCl\)
3 \(A{l^{3 + }} + 3C{l^ - }\)
4 \(A{l_2}{O_3} + 6HCl\)
Explanation:
Correct option is B) Aluminium chloride exists as dimer \(\left( Al _2 Cl _6\right)\) in a solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives \(\left[ Al ( OH )_6\right]^{3-}+3 HCl\) \(Al _2 Cl _6+12 H _2 O \rightarrow 2\left[ Al ( OH )_6\right]^{3-}+12 HCl\) Thus aluminum chloride is hydrolyzed to form the complex ion containing s hydroxide groups.
11. THE P-BLOCK ELEMENTS
189636
The hardest substance amongst the following is
1 \(B{e_2}C\)
2 Graphite
3 Titanium
4 \({B_4}C\)
Explanation:
(d)\({B_4}C\) is the hardest substance along with diamond.
11. THE P-BLOCK ELEMENTS
189619
The type of hybridisation of boron in diborane is
1 \(sp\) - hybridisation
2 \(s{p^2}{\rm{ - }}\) hybridisation
3 \(s{p^3}{\rm{ - }}\) hybridisation
4 \(s{p^3}{d^2}{\rm{ - }}\) hybridisation
Explanation:
Boron has three valence electrons, so it is supposed to make \(3\) bonds in a molecule with hybridization, \(sp ^2\) as only s and two \(p\) orbitals are used in hybridization and last \(p\) orbitalis vacant. But diborane, \(B _2 H _6\) contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry. Hence, each Boron atom is \(sp ^3\) - hybridized.
11. THE P-BLOCK ELEMENTS
189620
In the reaction \({B_2}{O_3} + C + C{l_2} \to A + CO.\) The \(A\) is
1 \(BC{l_3}\)
2 \(BC{l_2}\)
3 \({B_2}C{l_2}\)
4 \(CC{l_2}\)
Explanation:
(a) \({B_2}{O_3} + 3C + 3C{l_2} \to 2BC{l_3} + 3CO\) \(BC{l_3}\) is obtained by passing chlorine over the heated mixture of \({B_2}{O_3}\) and powdered charcoal.
189635
Aluminium chloride exists as dimer, \(A{l_2}C{l_6}\) in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives
1 \({[Al{(OH)_6}]^{3 - }} + 3HCl\)
2 \({[Al{({H_2}O)_6}]^{3 + }} + 3HCl\)
3 \(A{l^{3 + }} + 3C{l^ - }\)
4 \(A{l_2}{O_3} + 6HCl\)
Explanation:
Correct option is B) Aluminium chloride exists as dimer \(\left( Al _2 Cl _6\right)\) in a solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives \(\left[ Al ( OH )_6\right]^{3-}+3 HCl\) \(Al _2 Cl _6+12 H _2 O \rightarrow 2\left[ Al ( OH )_6\right]^{3-}+12 HCl\) Thus aluminum chloride is hydrolyzed to form the complex ion containing s hydroxide groups.
11. THE P-BLOCK ELEMENTS
189636
The hardest substance amongst the following is
1 \(B{e_2}C\)
2 Graphite
3 Titanium
4 \({B_4}C\)
Explanation:
(d)\({B_4}C\) is the hardest substance along with diamond.
11. THE P-BLOCK ELEMENTS
189619
The type of hybridisation of boron in diborane is
1 \(sp\) - hybridisation
2 \(s{p^2}{\rm{ - }}\) hybridisation
3 \(s{p^3}{\rm{ - }}\) hybridisation
4 \(s{p^3}{d^2}{\rm{ - }}\) hybridisation
Explanation:
Boron has three valence electrons, so it is supposed to make \(3\) bonds in a molecule with hybridization, \(sp ^2\) as only s and two \(p\) orbitals are used in hybridization and last \(p\) orbitalis vacant. But diborane, \(B _2 H _6\) contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry. Hence, each Boron atom is \(sp ^3\) - hybridized.
11. THE P-BLOCK ELEMENTS
189620
In the reaction \({B_2}{O_3} + C + C{l_2} \to A + CO.\) The \(A\) is
1 \(BC{l_3}\)
2 \(BC{l_2}\)
3 \({B_2}C{l_2}\)
4 \(CC{l_2}\)
Explanation:
(a) \({B_2}{O_3} + 3C + 3C{l_2} \to 2BC{l_3} + 3CO\) \(BC{l_3}\) is obtained by passing chlorine over the heated mixture of \({B_2}{O_3}\) and powdered charcoal.
189635
Aluminium chloride exists as dimer, \(A{l_2}C{l_6}\) in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives
1 \({[Al{(OH)_6}]^{3 - }} + 3HCl\)
2 \({[Al{({H_2}O)_6}]^{3 + }} + 3HCl\)
3 \(A{l^{3 + }} + 3C{l^ - }\)
4 \(A{l_2}{O_3} + 6HCl\)
Explanation:
Correct option is B) Aluminium chloride exists as dimer \(\left( Al _2 Cl _6\right)\) in a solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives \(\left[ Al ( OH )_6\right]^{3-}+3 HCl\) \(Al _2 Cl _6+12 H _2 O \rightarrow 2\left[ Al ( OH )_6\right]^{3-}+12 HCl\) Thus aluminum chloride is hydrolyzed to form the complex ion containing s hydroxide groups.
11. THE P-BLOCK ELEMENTS
189636
The hardest substance amongst the following is
1 \(B{e_2}C\)
2 Graphite
3 Titanium
4 \({B_4}C\)
Explanation:
(d)\({B_4}C\) is the hardest substance along with diamond.
11. THE P-BLOCK ELEMENTS
189619
The type of hybridisation of boron in diborane is
1 \(sp\) - hybridisation
2 \(s{p^2}{\rm{ - }}\) hybridisation
3 \(s{p^3}{\rm{ - }}\) hybridisation
4 \(s{p^3}{d^2}{\rm{ - }}\) hybridisation
Explanation:
Boron has three valence electrons, so it is supposed to make \(3\) bonds in a molecule with hybridization, \(sp ^2\) as only s and two \(p\) orbitals are used in hybridization and last \(p\) orbitalis vacant. But diborane, \(B _2 H _6\) contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry. Hence, each Boron atom is \(sp ^3\) - hybridized.
11. THE P-BLOCK ELEMENTS
189620
In the reaction \({B_2}{O_3} + C + C{l_2} \to A + CO.\) The \(A\) is
1 \(BC{l_3}\)
2 \(BC{l_2}\)
3 \({B_2}C{l_2}\)
4 \(CC{l_2}\)
Explanation:
(a) \({B_2}{O_3} + 3C + 3C{l_2} \to 2BC{l_3} + 3CO\) \(BC{l_3}\) is obtained by passing chlorine over the heated mixture of \({B_2}{O_3}\) and powdered charcoal.
189635
Aluminium chloride exists as dimer, \(A{l_2}C{l_6}\) in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives
1 \({[Al{(OH)_6}]^{3 - }} + 3HCl\)
2 \({[Al{({H_2}O)_6}]^{3 + }} + 3HCl\)
3 \(A{l^{3 + }} + 3C{l^ - }\)
4 \(A{l_2}{O_3} + 6HCl\)
Explanation:
Correct option is B) Aluminium chloride exists as dimer \(\left( Al _2 Cl _6\right)\) in a solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives \(\left[ Al ( OH )_6\right]^{3-}+3 HCl\) \(Al _2 Cl _6+12 H _2 O \rightarrow 2\left[ Al ( OH )_6\right]^{3-}+12 HCl\) Thus aluminum chloride is hydrolyzed to form the complex ion containing s hydroxide groups.
11. THE P-BLOCK ELEMENTS
189636
The hardest substance amongst the following is
1 \(B{e_2}C\)
2 Graphite
3 Titanium
4 \({B_4}C\)
Explanation:
(d)\({B_4}C\) is the hardest substance along with diamond.
11. THE P-BLOCK ELEMENTS
189619
The type of hybridisation of boron in diborane is
1 \(sp\) - hybridisation
2 \(s{p^2}{\rm{ - }}\) hybridisation
3 \(s{p^3}{\rm{ - }}\) hybridisation
4 \(s{p^3}{d^2}{\rm{ - }}\) hybridisation
Explanation:
Boron has three valence electrons, so it is supposed to make \(3\) bonds in a molecule with hybridization, \(sp ^2\) as only s and two \(p\) orbitals are used in hybridization and last \(p\) orbitalis vacant. But diborane, \(B _2 H _6\) contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry. Hence, each Boron atom is \(sp ^3\) - hybridized.
11. THE P-BLOCK ELEMENTS
189620
In the reaction \({B_2}{O_3} + C + C{l_2} \to A + CO.\) The \(A\) is
1 \(BC{l_3}\)
2 \(BC{l_2}\)
3 \({B_2}C{l_2}\)
4 \(CC{l_2}\)
Explanation:
(a) \({B_2}{O_3} + 3C + 3C{l_2} \to 2BC{l_3} + 3CO\) \(BC{l_3}\) is obtained by passing chlorine over the heated mixture of \({B_2}{O_3}\) and powdered charcoal.
