231.
Which group contains coloured ions out of the following?
$$\eqalign{
& \left( {\text{i}} \right)C{u^ + } \cr
& \left( {{\text{ii}}} \right)T{i^{4 + }} \cr
& \left( {{\text{iii}}} \right)C{o^{2 + }} \cr
& \left( {{\text{iv}}} \right)F{e^{2 + }} \cr} $$
A
(i), (ii), (iii), (iv)
B
(iii), (iv)
C
(ii), (iii)
D
(i), (ii)
Answer :
(iii), (iv)
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Colour of transition elements are due to presence of unpaired electrons.
232.
Cinnabar is an ore of
A
$$Hg$$
B
$$Cu$$
C
$$Pb$$
D
$$Zn$$
Answer :
$$Hg$$
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Cinnabar is an ore of mercury which have formula $$HgS.$$
233.
Among the following, the compound that is both paramagnetic and coloured, is
A
$$KMn{O_4}$$
B
$$Cu{F_2}$$
C
$${K_2}C{r_2}{O_7}$$
D
$${\text{All are coloured}}$$
Answer :
$$Cu{F_2}$$
234.
Which of the following may be considered to be an organometallic compound?
A
\[\text{Nickel tetracarbonyl}\]
B
\[\text{Chlorophyll}\]
C
$${K_3}\left[ {Fe{{\left( {{C_2}{O_4}} \right)}_3}} \right]$$
D
$$\left[ {Co{{\left( {en} \right)}_3}} \right]C{l_3}$$
Answer :
\[\text{Chlorophyll}\]
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Chlorophyll may be considered to be an organometallic compound because it contain metal carbon bond.
235.
Actinoids in general show more oxidation states than the lanthanoids. The main reason for this is
A
higher energy difference between $$5f$$ and $$6d$$ orbitals than between $$4f$$ and $$5d$$ orbitals
B
lower energy difference between $$5f$$ and $$6d$$ orbitals than between $$4f$$ and $$5d$$ orbitals
C
higher reactivity of actinoids than lanthanoids
D
actinoids are more basic than lanthanoids
Answer :
lower energy difference between $$5f$$ and $$6d$$ orbitals than between $$4f$$ and $$5d$$ orbitals
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No explanation is given for this question. Let's discuss the answer together.
236.
One mole of acidified $${K_2}C{r_2}{O_7}$$ on reaction with excess $$KI$$ will liberate ______ moles$$(s)$$ of $${I_2}.$$
A
3
B
1
C
7
D
2
Answer :
3
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237.
When $$KMn{O_4}$$ solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because
A
$$C{O_2}$$ is formed as the product
B
reaction is exothermic
C
$$MnO_4^ - $$ catalyses the reaction
D
$$M{n^{2 + }}$$ acts as autocatalyst
Answer :
$$M{n^{2 + }}$$ acts as autocatalyst
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238.
Which one of the following species is stable in aqueous solution ?
A
$$C{r^{2 + }}$$
B
$$MnO_4^{2 - }$$
C
$$MnO_4^{3 - }$$
D
$$C{u^ + }$$
Answer :
$$MnO_4^{2 - }$$
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In $$MnO_4^{2 - }$$ manganese is in $$+ 6$$ oxidation state which is having highest stability. $$MnO_4^{2 - }$$ disproportionates in neutral or acidic solution.
$$3MnO_4^{2 - } + 4{H^ + } \to $$ $$2MnO_4^{2 - } + Mn{O_2} + 2{H_2}O$$
239.
In which of the following pairs are both the ions coloured in aqueous solution?
$$\left( {{\text{At}}{\text{. no}}{\text{.}}\,Sc = 21,Ti = 22,} \right.$$ $$\left. {Ni = 28,Cu = 29,Co = 27} \right)$$
A
$$N{i^{2 + }},T{i^{3 + }}$$
B
$$S{c^{3 + }},T{i^{3 + }}$$
C
$$S{c^{3 + }},C{o^{2 + }}$$
D
$$N{i^{2 + }},C{u^ + }$$
Answer :
$$N{i^{2 + }},T{i^{3 + }}$$
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$${}_{28}Ni = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^8},4{s^2}$$
$$N{i^{2 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^8}$$
$$3{d^8}$$
( 2 unpaired electrons)
$${}_{22}Ti = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^2},4{s^2}$$
$$T{i^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^1}$$ (1 unpaired electron)
$${}_{21}Sc = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^1},4{s^2}$$
$$S{c^{3 + }} = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}$$ (no unpaired electron)
$${}_{29}Cu = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^1}$$
$$C{u^ + } = 1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}}$$ (no unpaired electron)
Hence, in the above ions, $$N{i^{2 + }}$$ and $$T{i^{3 + }}$$ are coloured in aqueous solution due to the presence of unpaired electrons in d subshell.
240.
The lanthanide contraction is responsible for the fact that
A
$$Zr$$ and $$Yt$$ have about the same radius
B
$$Zr$$ and $$Nb$$ have similar oxidation state
C
$$Zr$$ and $$Hf$$ have about the same radius
D
$$Zr$$ and $$Zn$$ have the same oxidation state
Answer :
$$Zr$$ and $$Hf$$ have about the same radius
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The elements of second and third transition series resembles more in properties than the elements of first and second transition series. It is due to lanthanide contraction. So, due to lanthanide contraction $$Zr$$ and $$Hf$$ have the same radius and also known as twins.