189983
Consider the following route of reactions \({R_2}SiC{l_2} + Warer \to \left( A \right)\xrightarrow{{Polymarisation}}\left( B \right)\)Compound \((B)\) in above reaction is
1 Dimer silicone
2 Linear silicone
3 Cross linked silicone
4 Polymerisation of \((A)\) does not occur
Explanation:
11. THE P-BLOCK ELEMENTS
189984
The most basic oxide of elements in group \(14\) of the periodic table is
1 \(SiO_2\)
2 \(GrO\)
3 \(SnO_2\)
4 \(PbO\)
Explanation:
in a group oxide strength is depemd upon the atomic size so most size is \(Pb\) hence \(Pb\) oxide is most oxide. Lead also form an oxide \(Pb _3 O _4\) which is a mixed oxide of \(PbO\) and \(PbO 2\). Among the monoxides, \(CO\) is neutral, \(GeO\) is basic while \(SnO\) and \(PbO\) are amphoteric. In \(CO_2\), \(C\) is sp hybridised
11. THE P-BLOCK ELEMENTS
189985
\((Si_2O_5)_n^{2n-}\) anion is obtained when
1 no oxygen of a \(SiO_4^{4-}\) tetrahedron is shared with another \(SiO_4^{4-}\) tetrahedron
2 one oxygen of a \(SiO_4^{4-}\) tetrahedron is shared with another \(SiO_4^{4-}\)- tetrahedron
3 two oxygen of a \(SiO_4^{4-}\) tetrahedron are shared with another \(SiO_4^{4-}\) tetrahedron
4 three oxygen of a \(SiO_4^{4-}\) tetrahedron are shared with another \(SiO_4^{4-}\) tetrahedron
Explanation:
Total No. of oxygen atoms per silicon atom \(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+1=2.5\) \(\therefore\) Formula \(\mathrm{Si}_{2} \mathrm{O}_{5}^{2-}\)
11. THE P-BLOCK ELEMENTS
189987
The silicate anion in the mineral kinoite is a chain of three \(SiO_4\) tetrahedra that share corners with adjacent tetrahedra. The mineral also contains \(Ca^{2+}\) ions, \(Cu^{2+}\) ions, and water molecules in a \(1 : 1 : 1\) ratio mineral is represented as:
1 \(CaCuSi_3O_{10} · H_2O\)
2 \(CaCuSi_3O_{10} · 2H_2O\)
3 \(Ca_2Cu_2Si_3O_{10} · 2H_2O\)
4 none of these
Explanation:
As it shares corner with next so it forms linear structure with \(\left(\mathrm{Si} \mathrm{O}_{3}\right)_{\mathrm{n}}^{-2 \mathrm{n}} .\) Therefore, three unit will have total negative charge of eight and positive charge will be same, so formula is - \(\mathrm{Ca}_{2} \mathrm{Cu}_{2} \mathrm{Si}_{3} \mathrm{O}_{10} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)
189983
Consider the following route of reactions \({R_2}SiC{l_2} + Warer \to \left( A \right)\xrightarrow{{Polymarisation}}\left( B \right)\)Compound \((B)\) in above reaction is
1 Dimer silicone
2 Linear silicone
3 Cross linked silicone
4 Polymerisation of \((A)\) does not occur
Explanation:
11. THE P-BLOCK ELEMENTS
189984
The most basic oxide of elements in group \(14\) of the periodic table is
1 \(SiO_2\)
2 \(GrO\)
3 \(SnO_2\)
4 \(PbO\)
Explanation:
in a group oxide strength is depemd upon the atomic size so most size is \(Pb\) hence \(Pb\) oxide is most oxide. Lead also form an oxide \(Pb _3 O _4\) which is a mixed oxide of \(PbO\) and \(PbO 2\). Among the monoxides, \(CO\) is neutral, \(GeO\) is basic while \(SnO\) and \(PbO\) are amphoteric. In \(CO_2\), \(C\) is sp hybridised
11. THE P-BLOCK ELEMENTS
189985
\((Si_2O_5)_n^{2n-}\) anion is obtained when
1 no oxygen of a \(SiO_4^{4-}\) tetrahedron is shared with another \(SiO_4^{4-}\) tetrahedron
2 one oxygen of a \(SiO_4^{4-}\) tetrahedron is shared with another \(SiO_4^{4-}\)- tetrahedron
3 two oxygen of a \(SiO_4^{4-}\) tetrahedron are shared with another \(SiO_4^{4-}\) tetrahedron
4 three oxygen of a \(SiO_4^{4-}\) tetrahedron are shared with another \(SiO_4^{4-}\) tetrahedron
Explanation:
Total No. of oxygen atoms per silicon atom \(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+1=2.5\) \(\therefore\) Formula \(\mathrm{Si}_{2} \mathrm{O}_{5}^{2-}\)
11. THE P-BLOCK ELEMENTS
189987
The silicate anion in the mineral kinoite is a chain of three \(SiO_4\) tetrahedra that share corners with adjacent tetrahedra. The mineral also contains \(Ca^{2+}\) ions, \(Cu^{2+}\) ions, and water molecules in a \(1 : 1 : 1\) ratio mineral is represented as:
1 \(CaCuSi_3O_{10} · H_2O\)
2 \(CaCuSi_3O_{10} · 2H_2O\)
3 \(Ca_2Cu_2Si_3O_{10} · 2H_2O\)
4 none of these
Explanation:
As it shares corner with next so it forms linear structure with \(\left(\mathrm{Si} \mathrm{O}_{3}\right)_{\mathrm{n}}^{-2 \mathrm{n}} .\) Therefore, three unit will have total negative charge of eight and positive charge will be same, so formula is - \(\mathrm{Ca}_{2} \mathrm{Cu}_{2} \mathrm{Si}_{3} \mathrm{O}_{10} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)
189983
Consider the following route of reactions \({R_2}SiC{l_2} + Warer \to \left( A \right)\xrightarrow{{Polymarisation}}\left( B \right)\)Compound \((B)\) in above reaction is
1 Dimer silicone
2 Linear silicone
3 Cross linked silicone
4 Polymerisation of \((A)\) does not occur
Explanation:
11. THE P-BLOCK ELEMENTS
189984
The most basic oxide of elements in group \(14\) of the periodic table is
1 \(SiO_2\)
2 \(GrO\)
3 \(SnO_2\)
4 \(PbO\)
Explanation:
in a group oxide strength is depemd upon the atomic size so most size is \(Pb\) hence \(Pb\) oxide is most oxide. Lead also form an oxide \(Pb _3 O _4\) which is a mixed oxide of \(PbO\) and \(PbO 2\). Among the monoxides, \(CO\) is neutral, \(GeO\) is basic while \(SnO\) and \(PbO\) are amphoteric. In \(CO_2\), \(C\) is sp hybridised
11. THE P-BLOCK ELEMENTS
189985
\((Si_2O_5)_n^{2n-}\) anion is obtained when
1 no oxygen of a \(SiO_4^{4-}\) tetrahedron is shared with another \(SiO_4^{4-}\) tetrahedron
2 one oxygen of a \(SiO_4^{4-}\) tetrahedron is shared with another \(SiO_4^{4-}\)- tetrahedron
3 two oxygen of a \(SiO_4^{4-}\) tetrahedron are shared with another \(SiO_4^{4-}\) tetrahedron
4 three oxygen of a \(SiO_4^{4-}\) tetrahedron are shared with another \(SiO_4^{4-}\) tetrahedron
Explanation:
Total No. of oxygen atoms per silicon atom \(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+1=2.5\) \(\therefore\) Formula \(\mathrm{Si}_{2} \mathrm{O}_{5}^{2-}\)
11. THE P-BLOCK ELEMENTS
189987
The silicate anion in the mineral kinoite is a chain of three \(SiO_4\) tetrahedra that share corners with adjacent tetrahedra. The mineral also contains \(Ca^{2+}\) ions, \(Cu^{2+}\) ions, and water molecules in a \(1 : 1 : 1\) ratio mineral is represented as:
1 \(CaCuSi_3O_{10} · H_2O\)
2 \(CaCuSi_3O_{10} · 2H_2O\)
3 \(Ca_2Cu_2Si_3O_{10} · 2H_2O\)
4 none of these
Explanation:
As it shares corner with next so it forms linear structure with \(\left(\mathrm{Si} \mathrm{O}_{3}\right)_{\mathrm{n}}^{-2 \mathrm{n}} .\) Therefore, three unit will have total negative charge of eight and positive charge will be same, so formula is - \(\mathrm{Ca}_{2} \mathrm{Cu}_{2} \mathrm{Si}_{3} \mathrm{O}_{10} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)
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11. THE P-BLOCK ELEMENTS
189983
Consider the following route of reactions \({R_2}SiC{l_2} + Warer \to \left( A \right)\xrightarrow{{Polymarisation}}\left( B \right)\)Compound \((B)\) in above reaction is
1 Dimer silicone
2 Linear silicone
3 Cross linked silicone
4 Polymerisation of \((A)\) does not occur
Explanation:
11. THE P-BLOCK ELEMENTS
189984
The most basic oxide of elements in group \(14\) of the periodic table is
1 \(SiO_2\)
2 \(GrO\)
3 \(SnO_2\)
4 \(PbO\)
Explanation:
in a group oxide strength is depemd upon the atomic size so most size is \(Pb\) hence \(Pb\) oxide is most oxide. Lead also form an oxide \(Pb _3 O _4\) which is a mixed oxide of \(PbO\) and \(PbO 2\). Among the monoxides, \(CO\) is neutral, \(GeO\) is basic while \(SnO\) and \(PbO\) are amphoteric. In \(CO_2\), \(C\) is sp hybridised
11. THE P-BLOCK ELEMENTS
189985
\((Si_2O_5)_n^{2n-}\) anion is obtained when
1 no oxygen of a \(SiO_4^{4-}\) tetrahedron is shared with another \(SiO_4^{4-}\) tetrahedron
2 one oxygen of a \(SiO_4^{4-}\) tetrahedron is shared with another \(SiO_4^{4-}\)- tetrahedron
3 two oxygen of a \(SiO_4^{4-}\) tetrahedron are shared with another \(SiO_4^{4-}\) tetrahedron
4 three oxygen of a \(SiO_4^{4-}\) tetrahedron are shared with another \(SiO_4^{4-}\) tetrahedron
Explanation:
Total No. of oxygen atoms per silicon atom \(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+1=2.5\) \(\therefore\) Formula \(\mathrm{Si}_{2} \mathrm{O}_{5}^{2-}\)
11. THE P-BLOCK ELEMENTS
189987
The silicate anion in the mineral kinoite is a chain of three \(SiO_4\) tetrahedra that share corners with adjacent tetrahedra. The mineral also contains \(Ca^{2+}\) ions, \(Cu^{2+}\) ions, and water molecules in a \(1 : 1 : 1\) ratio mineral is represented as:
1 \(CaCuSi_3O_{10} · H_2O\)
2 \(CaCuSi_3O_{10} · 2H_2O\)
3 \(Ca_2Cu_2Si_3O_{10} · 2H_2O\)
4 none of these
Explanation:
As it shares corner with next so it forms linear structure with \(\left(\mathrm{Si} \mathrm{O}_{3}\right)_{\mathrm{n}}^{-2 \mathrm{n}} .\) Therefore, three unit will have total negative charge of eight and positive charge will be same, so formula is - \(\mathrm{Ca}_{2} \mathrm{Cu}_{2} \mathrm{Si}_{3} \mathrm{O}_{10} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)
