1.
Which of the following is amphoteric oxide?
$$M{n_2}{O_7},Cr{O_3},C{r_2}{O_3},CrO,{V_2}{O_5},{V_2}{O_4}$$
A.
$${V_2}{O_5},C{r_2}{O_3}$$
B.
$$M{n_2}{O_7},Cr{O_3}$$
C.
$$CrO,{V_2}{O_5}$$
D.
$${V_2}{O_5},{V_2}{O_4}$$
Answer :
$${V_2}{O_5},C{r_2}{O_3}$$
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2.
The outer electron configuration of $$Gd\left( {Atomic\,No.:64} \right)\,{\text{is}}:$$
A.
$$4{f^3}5{d^5}6{s^2}$$
B.
$$4{f^8}5{d^0}6{s^2}$$
C.
$$4{f^4}5{d^4}6{s^2}$$
D.
$$4{f^7}5{d^1}6{s^2}$$
Answer :
$$4{f^7}5{d^1}6{s^2}$$
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The configuration of $$Gd\,{\text{is}}\,\,4{f^7}5{d^1}6{s^2}.$$
3.
The stability of dihalides of $$Si,Ge,Sn$$ and $$Pb$$ increases steadily in the sequence
A.
$$Pb{X_2} < < Sn{X_2} < < Ge{X_2} < < Si{X_2}$$
B.
$$Ge{X_2} < < Si{X_2} < < Sn{X_2} < < Pb{X_2}$$
C.
$$Si{X_2} < < Ge{X_2} < < Pb{X_2} < < Sn{X_2}$$
D.
$$Si{X_2} < < Ge{X_2} < < Sn{X_2} < < Pb{X_2}$$
Answer :
$$Si{X_2} < < Ge{X_2} < < Sn{X_2} < < Pb{X_2}$$
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Reluctance of valence shell electrons to participate in bonding is called inert pair effect. The stability of lower oxidation state ( + 2 for group 14 element ) increases on going down the group. So the correct order is
$$Si{X_2} < < Ge{X_2} < < Sn{X_2} < < Pb{X_2}$$
4.
The common oxidation states of $$Ti$$ are
A.
+2, +3
B.
+3, +4
C.
3, -4
D.
+2, +3, +4
Answer :
+3, +4
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The most common oxidation state of titanium are +3 and +4.
5.
Calomel$$\left( {H{g_2}C{l_2}} \right)$$ on reaction with ammonium hydroxide gives
A.
$$HgO$$
B.
$$H{g_2}O$$
C.
$$N{H_2} - Hg - Hg - Cl$$
D.
$$Hg\,N{H_2}\,Cl$$
Answer :
$$Hg\,N{H_2}\,Cl$$
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$$H{g_2}C{l_2} + 2N{H_4}OH \to HgN{H_2}Cl + 2{H_2}O$$
6.
Which of the following statements is correct about stability of the complexes of lanthanoids?
A.
Stability of complexes increases as the size of lanthanoid decreases.
B.
Stability of complexes decreases as the size of lanthanoid decreases.
C.
Lanthanoids do not form complexes.
D.
All the complexes of lanthanoids have same stability.
Answer :
Stability of complexes increases as the size of lanthanoid decreases.
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7.
Which of the following statements is not correct?
A.
$$La{\left( {OH} \right)_3}$$ is less basic than $$Li{\left( {OH} \right)_3}$$
B.
In lanthanide series, ionic radius of $$L{n^{3 + }}$$ $$ion$$ decreases
C.
$$La$$ is actually an element of transition series rather lanthanide
D.
Atomic radius of $$Zr$$ and $$Hf$$ are same because of lanthanide contraction
Answer :
$$La{\left( {OH} \right)_3}$$ is less basic than $$Li{\left( {OH} \right)_3}$$
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$$L{a^{3 + }}$$ $$ions$$ larger than $$L{i^{3 + }}.$$ So, it easily gives $$O{H^ - }$$ $$ion$$ $$La{\left( {OH} \right)_3}$$ is more basic than $$Li{\left( {OH} \right)_3}.$$ In lanthanides the basic character of hydroxides
decreases as the ionic radius decreases.
8.
Which of the following ions will exhibit colour in aqueous solutions?
A.
$$L{a^{3 + }}\left( {Z = 57} \right)$$
B.
$$T{i^{3 + }}\left( {Z = 22} \right)$$
C.
$$L{u^{3 + }}\left( {Z = 71} \right)$$
D.
$$S{c^{3 + }}\left( {Z = 21} \right)$$
Answer :
$$T{i^{3 + }}\left( {Z = 22} \right)$$
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Key Idea Colour is obtained as a consequence of $$d-d$$ ( or $$f-f$$ ) transition, and for $$d-d$$ ( or $$f-f$$ ) transition,
presence of unpaired electrons is the necessary condition.
Electronic configuration of
$$L{a^{3 + }}\left( {Z = 57} \right) = \left[ {Xe} \right]4{f^0}5{d^0}6{s^0}$$ (no unpaired electron)
$$T{i^{3 + }}\left( {Z = 22} \right) = \left[ {Ar} \right]3{d^1}4{s^0}$$ (one unpaired electron)
$$L{u^{3 + }}\left( {Z = 71} \right) = \left[ {Xe} \right]4{f^{14}}5{d^0}6{s^0}$$ (no unpaired electron)
$$S{c^{3 + }}\left( {Z = 21} \right) = \left[ {Ar} \right]3{d^0}4{s^0}$$ (no unpaired electron)
Hence, due to the presence of unpaired electron in $$T{i^{3 + }},$$ it exhibit colour in aqueous solution.
9.
Which of the following arrangements does not represent the correct order of the property stated against it?
A.
$$Sc < Ti < Cr < Mn:$$ number of oxidation states
B.
$${V^{2 + }} < C{r^{2 + }} < M{n^{2 + }} < F{e^{2 + }}:$$ paramagnetic behaviour
C.
$$N{i^{2 + }} < C{o^{2 + }} < F{e^{2 + }} < M{n^{2 + }}:$$ ionic size
D.
$$C{o^{3 + }} < F{e^{3 + }} < C{r^{3 + }} < S{c^{3 + }}:$$ stability in aqueous solution
Answer :
$${V^{2 + }} < C{r^{2 + }} < M{n^{2 + }} < F{e^{2 + }}:$$ paramagnetic behaviour
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Number of unpaired electrons in $$F{e^{2 + }}$$ is less than $$M{n^{2 + }},$$ so $$F{e^{2 + }}$$ is less paramagnetic than $$M{n^{2 + }}.$$
10.
A metal $$M$$ and its compound can give the following observable changes in a consequence of reactions
$$\eqalign{
& M\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{HN{O_3}}^{{\rm{dilute}}}} \left[ \matrix{
{\rm{Colourless}} \hfill \cr
{\rm{solutions}} \hfill \cr} \right]\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{NaOH}^{{\rm{aqueous}}}} \left[ \matrix{
{\rm{White}} \hfill \cr
{\rm{precipitate}} \hfill \cr} \right] \cr
& \left[ \matrix{
{\rm{White}} \hfill \cr
{\rm{precipitate}} \hfill \cr} \right]\buildrel {{H_2}S} \over
\longrightarrow \left[ \matrix{
{\rm{Colourless}} \hfill \cr
{\rm{solutions}} \hfill \cr} \right]\mathrel{\mathop{\kern0pt\longleftarrow}
\limits_{NaOH\left( {aq.} \right)}^{{\rm{excess}}}} \cr} $$
A.
$$Mg$$
B.
$$Pb$$
C.
$$Zn$$
D.
$$Sn$$
Answer :
$$Zn$$
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\[\begin{align}
& \underset{\left( M \right)}{\mathop{Zn}}\,\xrightarrow{dil.\,\,HN{{O}_{3}}}\overset{\text{Brown}}{\mathop{\underset{\begin{smallmatrix}
\text{Colourless} \\
\text{solution}
\end{smallmatrix}}{\mathop{Zn{{\left( N{{O}_{3}} \right)}_{2}}}}\,\xrightarrow[aq.]{NaOH}}}\,\underset{\text{White ppt}\text{.}}{\mathop{Zn{{\left( OH \right)}_{2}}}}\, \\
& \xrightarrow[NaOH]{\text{Excess}}N{{a}_{2}}\underset{\text{Soluble}}{\mathop{\left[ Zn{{\left( OH \right)}_{4}} \right]}}\,\xrightarrow{{{H}_{2}}S}\underset{\text{White ppt}\text{.}}{\mathop{ZnS}}\, \\
\end{align}\]