Coordination Compounds Quiz-14
Dear Readers,
JEE Advanced is a high level exam that checks your concept by putting various types of questions. The questions are in a different format from JEE mains, herein the questions are : single correct multiple choice questions, multiple correct multiple choice questions, statement based questions and comprehension type questions..
Q1.
Statement 1: EDTA forms complexes with a large number of metal ions
Statement 2: It coordinates with 6 points of attachement to the metal, 4 O-atoms and two N atoms
Statement 1: EDTA forms complexes with a large number of metal ions
Statement 2: It coordinates with 6 points of attachement to the metal, 4 O-atoms and two N atoms
Solution
(a) EDTA binds to both Ca^(2+)and Mg^(2+),ie, cations with noble gas configuration as well as to transition metal ions. It is so efficient in binding metal ions that it is used to remove traces of metal from distilled water
(a) EDTA binds to both Ca^(2+)and Mg^(2+),ie, cations with noble gas configuration as well as to transition metal ions. It is so efficient in binding metal ions that it is used to remove traces of metal from distilled water
Q2.
Statement 1: Ions with more than five 3d electrons have usually slightly larger magnetic moment than calculated on the basis of μ=√(n(n+2))
Statement 2: The maximum number of unpaired d-electrons is five as in Mn^(2+)and Fe^(3+) and so, μ=5.92 for 5 unpaired electrons
Statement 1: Ions with more than five 3d electrons have usually slightly larger magnetic moment than calculated on the basis of μ=√(n(n+2))
Statement 2: The maximum number of unpaired d-electrons is five as in Mn^(2+)and Fe^(3+) and so, μ=5.92 for 5 unpaired electrons
Solution
(b) The slightly larger value of μ than expected from the formula μ=√(n(n+2)) is due to a small contribution from the orbital anguler momentum of the electrons to the magnetic moment
(b) The slightly larger value of μ than expected from the formula μ=√(n(n+2)) is due to a small contribution from the orbital anguler momentum of the electrons to the magnetic moment
Q3.
Statement 1: The conversion of an optically active compound into its enantiomer is called Walden inversion
Statement 2: A racemic mixture is optically inactive due to internal compensation
Statement 1: The conversion of an optically active compound into its enantiomer is called Walden inversion
Statement 2: A racemic mixture is optically inactive due to internal compensation
Solution
(c) Racemic mixture is optically inactive because the two enantiomers rotate the plane polarised light equally in opposite directions and cancel each others rotation. This phenomenon is called external compensation
(c) Racemic mixture is optically inactive because the two enantiomers rotate the plane polarised light equally in opposite directions and cancel each others rotation. This phenomenon is called external compensation
Q4.
Statement 1: Staggered form is less stable than the eclipsed form
Statement 2: The conformation in which the bond pairs of two central atoms are very far from one another is called staggered form
Solution
(d) The staggered form is more stable than the eclipsed form because the potential energy of staggered form in which the bond pairs of two carbons are far away from each other is minimum
(d) The staggered form is more stable than the eclipsed form because the potential energy of staggered form in which the bond pairs of two carbons are far away from each other is minimum
Q5.
Statement 1: d-d transition is not possible in [Sc(H_2 O)_6 ]^(3+)
Statement 2: [Ti(H_2 O)_6 ]^(4+) is coloured while [Sc(H_2 O)_6 ]^(3+) is colourless
Statement 1: d-d transition is not possible in [Sc(H_2 O)_6 ]^(3+)
Statement 2: [Ti(H_2 O)_6 ]^(4+) is coloured while [Sc(H_2 O)_6 ]^(3+) is colourless
Solution
(c) Both [Ti(H_2 O)_6 ]^(4+)and [Sc(H_2 O)_6 ]^(3+) are colourless due to absence of free electrons in 3d subshell
(c) Both [Ti(H_2 O)_6 ]^(4+)and [Sc(H_2 O)_6 ]^(3+) are colourless due to absence of free electrons in 3d subshell
Q6. Statement 1: CH_3-CH(CH_3)-CH2-CO-OH is 3-methyl butanoic acid
Statement 2: In poly functional group, the substituent should be given lower number than the principal functional group
Solution
(c) The functional group is –COOH, the numbering is done from RHS to give minimum number to carbon atom bearing the functional group. The given compound is a derivative of butane. The substituent is the methyl group. So, the above compound is 3-methyl butanoic acid
(c) The functional group is –COOH, the numbering is done from RHS to give minimum number to carbon atom bearing the functional group. The given compound is a derivative of butane. The substituent is the methyl group. So, the above compound is 3-methyl butanoic acid
Q7.
Statement 1: Geometrical isomerism is also called C is- trans isomerism.
Statement 2: Tetrahedral complexes shows geometrical isomerism.
Statement 1: Geometrical isomerism is also called C is- trans isomerism.
Statement 2: Tetrahedral complexes shows geometrical isomerism.
Solution
(c) Square planar complexes having dsp^2 hybridisation shows geometrical isomerism.
(c) Square planar complexes having dsp^2 hybridisation shows geometrical isomerism.
Q8.
Statement 1: In keto-enol tautomerism of dicarbonyl compounds, the enol form is preferred in contrast to the keto-form
Statement 2: The enol form is more stable due to resonance
Statement 1: In keto-enol tautomerism of dicarbonyl compounds, the enol form is preferred in contrast to the keto-form
Statement 2: The enol form is more stable due to resonance
Solution
(a) Resonance stabilisation of enol form can be shown as
(a) Resonance stabilisation of enol form can be shown as
Q9.
Statement 1: [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless.
Statement 2: d-d transition is not possible in [Sc(H2O)6]3+.
Statement 1: [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless.
Statement 2: d-d transition is not possible in [Sc(H2O)6]3+.
Solution
(a) [Sc[H_2 O_6 ]^(3+)] has no unpaired electrons in its d-subshell and thus d-d transition is not possible whereas [Ti(H_2 O_6 )]^(3+) has one unpaired electron in its d-subshell which gives rise to d-d transition to impart colour.
(a) [Sc[H_2 O_6 ]^(3+)] has no unpaired electrons in its d-subshell and thus d-d transition is not possible whereas [Ti(H_2 O_6 )]^(3+) has one unpaired electron in its d-subshell which gives rise to d-d transition to impart colour.
Q10.
Statement 1: A hydroxy group directly attached to a carbonyl group constitutes a carboxyl group
Statement 2: Ester is a family of carboxylic acid derivatives in which –OH group is altered by other group
Statement 1: A hydroxy group directly attached to a carbonyl group constitutes a carboxyl group
Statement 2: Ester is a family of carboxylic acid derivatives in which –OH group is altered by other group
Solution
(b) If the acidic hydrogen of carboxylic acid is replaced by an aryl or alkyl group, the resulting structure is a carboxylate ester
(b) If the acidic hydrogen of carboxylic acid is replaced by an aryl or alkyl group, the resulting structure is a carboxylate ester
