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CHXI10:THE S-BLOCK ELEMENTS

315935 Amongst $LiCl, RbCl, {BeCl}_{2}$ and ${MgCl}_{2}$ the compounds with the greatest and least ionic character respectively are

1

LiCl

and

RbCl

2

{MgCl}_{2}

and

{BeCl}_{2}

3

RbCl

and

{BeCl}_{2}

4

RbCl

and

{MgCl}_{2}

Explanation:

According to Fajan's rule,
Size of cation $\propto$ Ionic character.
Recall that size of metal (ion) increases while going down the gorup, and decreases on crossing the periods from left to right. Thus ${Rb}^{+}$ (group 1) is largest and ${Be}^{2 +}$ (group II) is smallest in size. hence $RbCl$ has greatest, and ${BeCl}_{2}$ has lowest ionic character.

CHXI10:THE S-BLOCK ELEMENTS

315936 $HCl$ was passed through solution of ${CaCl}_{2}$ , ${MgCl}_{2}$ and $NaCl$ . Which of the following compound(s) crystallise (s)?

1 Only

NaCl

2 Only

{MgCl}_{2}

3

NaCl, {MgCl}_{2}

and

{CaCl}_{2}

4 Both

{MgCl}_{2}

and

{CaCl}_{2}

Explanation:

When $HCl$ is passed through the mixture ${Cl}^{-}$ ion concentration increase.Hence ionic product $[{Na}^{+}] [{Cl}^{-}]$ becomes more than solubility product.
So, $NaCl$ will precipitated out
Filtrate $\overset{HCl gas passed}{⟶}$ pure $NaCl$ precipitation (Common ion effect)

NEET - 2020

CHXI10:THE S-BLOCK ELEMENTS

315937 The salt which responds to dilute and concentrated $H_{2} {SO}_{4}$ is

1

Ba {({NO}_{3})}_{2}

2

{Na}_{3} {PO}_{4}

3

{CaF}_{2}

4

{Na}_{2} {So}_{4}

Explanation:

$Ba {({NO}_{3})}_{2}$ responds to both dilute and concentrated $H_{2} {SO}_{4}$ .
$B a {(N O_{3})}_{2} + H_{2} S O_{4} (d i l) \to B a S O_{4 (s)} ↓$
$+ 2 H^{(a q)} {+ 2 N O}_{3}^{- (a q)}$
$B a {(N O_{3})}_{2} + H_{2} S O_{4} (c o n c) \to \underset{(w h i t e p p t)}{B a S O_{4}} + 2 H N O_{3}$
$o r {(C a S O_{4})}_{2} \cdot H_{2} O$

KCET - 2015

CHXI10:THE S-BLOCK ELEMENTS

315935 Amongst $LiCl, RbCl, {BeCl}_{2}$ and ${MgCl}_{2}$ the compounds with the greatest and least ionic character respectively are

1

LiCl

and

RbCl

2

{MgCl}_{2}

and

{BeCl}_{2}

3

RbCl

and

{BeCl}_{2}

4

RbCl

and

{MgCl}_{2}

Explanation:

According to Fajan's rule,
Size of cation $\propto$ Ionic character.
Recall that size of metal (ion) increases while going down the gorup, and decreases on crossing the periods from left to right. Thus ${Rb}^{+}$ (group 1) is largest and ${Be}^{2 +}$ (group II) is smallest in size. hence $RbCl$ has greatest, and ${BeCl}_{2}$ has lowest ionic character.

CHXI10:THE S-BLOCK ELEMENTS

315936 $HCl$ was passed through solution of ${CaCl}_{2}$ , ${MgCl}_{2}$ and $NaCl$ . Which of the following compound(s) crystallise (s)?

1 Only

NaCl

2 Only

{MgCl}_{2}

3

NaCl, {MgCl}_{2}

and

{CaCl}_{2}

4 Both

{MgCl}_{2}

and

{CaCl}_{2}

Explanation:

When $HCl$ is passed through the mixture ${Cl}^{-}$ ion concentration increase.Hence ionic product $[{Na}^{+}] [{Cl}^{-}]$ becomes more than solubility product.
So, $NaCl$ will precipitated out
Filtrate $\overset{HCl gas passed}{⟶}$ pure $NaCl$ precipitation (Common ion effect)

NEET - 2020

CHXI10:THE S-BLOCK ELEMENTS

315937 The salt which responds to dilute and concentrated $H_{2} {SO}_{4}$ is

1

Ba {({NO}_{3})}_{2}

2

{Na}_{3} {PO}_{4}

3

{CaF}_{2}

4

{Na}_{2} {So}_{4}

Explanation:

$Ba {({NO}_{3})}_{2}$ responds to both dilute and concentrated $H_{2} {SO}_{4}$ .
$B a {(N O_{3})}_{2} + H_{2} S O_{4} (d i l) \to B a S O_{4 (s)} ↓$
$+ 2 H^{(a q)} {+ 2 N O}_{3}^{- (a q)}$
$B a {(N O_{3})}_{2} + H_{2} S O_{4} (c o n c) \to \underset{(w h i t e p p t)}{B a S O_{4}} + 2 H N O_{3}$
$o r {(C a S O_{4})}_{2} \cdot H_{2} O$

KCET - 2015

CHXI10:THE S-BLOCK ELEMENTS

315935 Amongst $LiCl, RbCl, {BeCl}_{2}$ and ${MgCl}_{2}$ the compounds with the greatest and least ionic character respectively are

1

LiCl

and

RbCl

2

{MgCl}_{2}

and

{BeCl}_{2}

3

RbCl

and

{BeCl}_{2}

4

RbCl

and

{MgCl}_{2}

Explanation:

According to Fajan's rule,
Size of cation $\propto$ Ionic character.
Recall that size of metal (ion) increases while going down the gorup, and decreases on crossing the periods from left to right. Thus ${Rb}^{+}$ (group 1) is largest and ${Be}^{2 +}$ (group II) is smallest in size. hence $RbCl$ has greatest, and ${BeCl}_{2}$ has lowest ionic character.

CHXI10:THE S-BLOCK ELEMENTS

315936 $HCl$ was passed through solution of ${CaCl}_{2}$ , ${MgCl}_{2}$ and $NaCl$ . Which of the following compound(s) crystallise (s)?

1 Only

NaCl

2 Only

{MgCl}_{2}

3

NaCl, {MgCl}_{2}

and

{CaCl}_{2}

4 Both

{MgCl}_{2}

and

{CaCl}_{2}

Explanation:

When $HCl$ is passed through the mixture ${Cl}^{-}$ ion concentration increase.Hence ionic product $[{Na}^{+}] [{Cl}^{-}]$ becomes more than solubility product.
So, $NaCl$ will precipitated out
Filtrate $\overset{HCl gas passed}{⟶}$ pure $NaCl$ precipitation (Common ion effect)

NEET - 2020

CHXI10:THE S-BLOCK ELEMENTS

315937 The salt which responds to dilute and concentrated $H_{2} {SO}_{4}$ is

1

Ba {({NO}_{3})}_{2}

2

{Na}_{3} {PO}_{4}

3

{CaF}_{2}

4

{Na}_{2} {So}_{4}

Explanation:

$Ba {({NO}_{3})}_{2}$ responds to both dilute and concentrated $H_{2} {SO}_{4}$ .
$B a {(N O_{3})}_{2} + H_{2} S O_{4} (d i l) \to B a S O_{4 (s)} ↓$
$+ 2 H^{(a q)} {+ 2 N O}_{3}^{- (a q)}$
$B a {(N O_{3})}_{2} + H_{2} S O_{4} (c o n c) \to \underset{(w h i t e p p t)}{B a S O_{4}} + 2 H N O_{3}$
$o r {(C a S O_{4})}_{2} \cdot H_{2} O$

KCET - 2015

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