NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI11:THE P-BLOCK ELEMENTS
316444
Which among the following pairs of halogen forms the interhalogen compound of the type \(\mathrm{XX}_{7}^{\prime}\) ?
1 \(\mathrm{Cl}\) and \(\mathrm{F}\)
2 I and \(F\)
3 I and \(\mathrm{Cl}\)
4 \({\rm{Br}}\,\,{\rm{and}}\,\,{\rm{F}}\)
Explanation:
Iodine (I) and fluorine (F) pairs of halogen forms the interhalogen compound of the type \(\mathrm{XX}_{7}{ }_{7}\), that is iodine heptafluoride with the chemical formula \(\mathrm{IF}_{7}\). It has an unusual pentagonal bipyramidal structure, as predicted by VSEPR theory
MHTCET - 2020
CHXI11:THE P-BLOCK ELEMENTS
316462
The total number of lone pairs of electrons in an \({\mathrm{\mathrm{IF}_{7}}}\) molecule is ____ .
1 0
2 7
3 42
4 21
Explanation:
the structure of \({\mathrm{\mathrm{IF}_{7}}}\) is as follows: There is no lone pair of electrons on central \({\text{I}}\) atom. However, each of seven \({\mathrm{F}}\) atoms contains 3 lone pairs of electrons. Hence, total number of lone pairs of electrons in an \({\rm{I}}{{\rm{F}}_{\rm{7}}}\) molecule is 21.
CHXI11:THE P-BLOCK ELEMENTS
316463
\(\mathrm{IBr}_{7}\) cannot exist but \(\mathrm{IF}_{7}\) exists. This fact can be explained on the basis of:
1 Ionisation potential
2 Electron affinities
3 Ratio of radii of atoms
4 Reducing abilities
Explanation:
Around 'I', seven small ' \(\mathrm{F}\) ' atoms can be accommodated where as big ' \(\mathrm{Br}\) ' atoms are not easily accommodated. On the basis of electronegativity also it can be explained. ' \(\mathrm{F}\) ' due to high e.n, can form \(\mathrm{IF}_{7}\) where as \(\mathrm{IBr}_{7}\) is not possible due to less electronegativity of \(\mathrm{Br}\).
316444
Which among the following pairs of halogen forms the interhalogen compound of the type \(\mathrm{XX}_{7}^{\prime}\) ?
1 \(\mathrm{Cl}\) and \(\mathrm{F}\)
2 I and \(F\)
3 I and \(\mathrm{Cl}\)
4 \({\rm{Br}}\,\,{\rm{and}}\,\,{\rm{F}}\)
Explanation:
Iodine (I) and fluorine (F) pairs of halogen forms the interhalogen compound of the type \(\mathrm{XX}_{7}{ }_{7}\), that is iodine heptafluoride with the chemical formula \(\mathrm{IF}_{7}\). It has an unusual pentagonal bipyramidal structure, as predicted by VSEPR theory
MHTCET - 2020
CHXI11:THE P-BLOCK ELEMENTS
316462
The total number of lone pairs of electrons in an \({\mathrm{\mathrm{IF}_{7}}}\) molecule is ____ .
1 0
2 7
3 42
4 21
Explanation:
the structure of \({\mathrm{\mathrm{IF}_{7}}}\) is as follows: There is no lone pair of electrons on central \({\text{I}}\) atom. However, each of seven \({\mathrm{F}}\) atoms contains 3 lone pairs of electrons. Hence, total number of lone pairs of electrons in an \({\rm{I}}{{\rm{F}}_{\rm{7}}}\) molecule is 21.
CHXI11:THE P-BLOCK ELEMENTS
316463
\(\mathrm{IBr}_{7}\) cannot exist but \(\mathrm{IF}_{7}\) exists. This fact can be explained on the basis of:
1 Ionisation potential
2 Electron affinities
3 Ratio of radii of atoms
4 Reducing abilities
Explanation:
Around 'I', seven small ' \(\mathrm{F}\) ' atoms can be accommodated where as big ' \(\mathrm{Br}\) ' atoms are not easily accommodated. On the basis of electronegativity also it can be explained. ' \(\mathrm{F}\) ' due to high e.n, can form \(\mathrm{IF}_{7}\) where as \(\mathrm{IBr}_{7}\) is not possible due to less electronegativity of \(\mathrm{Br}\).
316444
Which among the following pairs of halogen forms the interhalogen compound of the type \(\mathrm{XX}_{7}^{\prime}\) ?
1 \(\mathrm{Cl}\) and \(\mathrm{F}\)
2 I and \(F\)
3 I and \(\mathrm{Cl}\)
4 \({\rm{Br}}\,\,{\rm{and}}\,\,{\rm{F}}\)
Explanation:
Iodine (I) and fluorine (F) pairs of halogen forms the interhalogen compound of the type \(\mathrm{XX}_{7}{ }_{7}\), that is iodine heptafluoride with the chemical formula \(\mathrm{IF}_{7}\). It has an unusual pentagonal bipyramidal structure, as predicted by VSEPR theory
MHTCET - 2020
CHXI11:THE P-BLOCK ELEMENTS
316462
The total number of lone pairs of electrons in an \({\mathrm{\mathrm{IF}_{7}}}\) molecule is ____ .
1 0
2 7
3 42
4 21
Explanation:
the structure of \({\mathrm{\mathrm{IF}_{7}}}\) is as follows: There is no lone pair of electrons on central \({\text{I}}\) atom. However, each of seven \({\mathrm{F}}\) atoms contains 3 lone pairs of electrons. Hence, total number of lone pairs of electrons in an \({\rm{I}}{{\rm{F}}_{\rm{7}}}\) molecule is 21.
CHXI11:THE P-BLOCK ELEMENTS
316463
\(\mathrm{IBr}_{7}\) cannot exist but \(\mathrm{IF}_{7}\) exists. This fact can be explained on the basis of:
1 Ionisation potential
2 Electron affinities
3 Ratio of radii of atoms
4 Reducing abilities
Explanation:
Around 'I', seven small ' \(\mathrm{F}\) ' atoms can be accommodated where as big ' \(\mathrm{Br}\) ' atoms are not easily accommodated. On the basis of electronegativity also it can be explained. ' \(\mathrm{F}\) ' due to high e.n, can form \(\mathrm{IF}_{7}\) where as \(\mathrm{IBr}_{7}\) is not possible due to less electronegativity of \(\mathrm{Br}\).
