QBManager

Classification of Elements and Periodicity in Properties

89368 Assertion: Both $Be$ and $Al$ can form complexes such as ${BeF}_{4}^{2 -}$ and ${AlF}_{6}^{3 -}$ respectively, ${BeF}_{6}^{3 -}$ is not formed.
Reason: In case of $B e$ , no vacant d-orbital's are present in its outermost shell.

1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

$Be$ and $Al$ can form complexes such as ${BeF}_{4}^{2 -}$ and ${AlF}_{6}^{3 -}$ respectively, ${BeF}_{6}^{3 -}$ is not formed due to the absence of vacant d-orbital is not enough orbitals to accommodate extra electrons.
Hence, both asseration and reasons are correct and reason is the correct explanation of the assertion.

AIIMS-2015

Classification of Elements and Periodicity in Properties

89379 Lattice energy of an ionic compound depends upon

1 charge on the ion only

2 size of the ion only

3 charge on the ion and size of the ion

4 packing of its ions only

Explanation:

From the Born-Lande equation-
$U = \frac{N_{A} {Mz}^{+} z^{-} e^{2}}{4 π ϵ_{0} r} (1 - \frac{1}{n})$
Where, $U =$ Lattice energy of crystal
$N_{A} =$ Avogadro's number
$M =$ Molecular weight
$z^{+}, z^{-} =$ Charge of cation and anion
$r =$ Radius of the atom
$n =$ Exponent
From the above equation it is clear that lattice energy depends upon charge on the ion and size of the ion.

AP - EAMCET(MEDICAL) - 2009

Classification of Elements and Periodicity in Properties

89395 What is the correct relationship between the pH of isomolar solutions of sodium oxide, ${Na}_{2} O$ $({pH}_{1})$ , sodium sulphide, ${Na}_{2} S ({pH}_{2})$ , sodium selenide, ${Na}_{2} Se ({pH}_{3})$ and sodium telluride ${Na}_{2} T e ({pH}_{4})$ ?

1

{pH}_{1} > {pH}_{2} > {pH}_{3} > {pH}_{4}

2

{pH}_{1} > {pH}_{2} = {pH}_{3} > {pH}_{4}

3

{pH}_{1} > {pH}_{2} < {pH}_{3} < {pH}_{4}

4

{pH}_{1} > {pH}_{2} < {pH}_{3} \approx {pH}_{4}

Explanation:

The basicity of these oxides decreases, when placed in water, the basicity of solution also decreases.
${Na}_{2} O + 2 H_{2} O ⟶ 2 NaOH + H_{2} O$
${Na}_{2} S + 2 H_{2} O ⟶ 2 NaOH + H_{2} S$
${Na}_{2} Se + 2 H_{2} O ⟶ 2 NaOH + H_{2} Se$
${Na}_{2} Te + 2 H_{2} O ⟶ 2 NaOH + H_{2} Te$
Hence, the order of neutralisation of $NaOH$ is $H_{2} Te >$ $H_{2} Se > H_{2} S > H_{2} O$ .
Their aqueous solution have the following order of basic character due to neutralization of $NaOH$ with ${Na}_{2} O > {Na}_{2} S > {Na}_{2} Se > {Na}_{2} Te$ or $({pH}_{1}) > ({pH}_{2}) > ({pH}_{3}) > ({PH}_{4})$

NEET-2005

Classification of Elements and Periodicity in Properties

89401 Among the following elements that has the lowest first ionization potential is

1 nitrogen

2 oxygen

3 fluorine

4 neon

Explanation:

$_{7} N = 1 s^{2} 2 s^{2} 2 p^{3}$
$\begin{aligned} _{8} O = 1 s^{2} 2 s^{2} 2 p^{4} \\ _{9} F = 1 s^{2} 2 s^{2} 2 p^{5} \\ _{10} Ne = 1 s^{2} 2 s^{2} 2 p^{6} \end{aligned}$
Nitrogen has half-filled orbital and Ne has filled orbital it show stable configuration but Oxygen has less stable configuration and less nuclear charge.

A.P.EAMCET 1994

Classification of Elements and Periodicity in Properties

89368 Assertion: Both $Be$ and $Al$ can form complexes such as ${BeF}_{4}^{2 -}$ and ${AlF}_{6}^{3 -}$ respectively, ${BeF}_{6}^{3 -}$ is not formed.
Reason: In case of $B e$ , no vacant d-orbital's are present in its outermost shell.

1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

$Be$ and $Al$ can form complexes such as ${BeF}_{4}^{2 -}$ and ${AlF}_{6}^{3 -}$ respectively, ${BeF}_{6}^{3 -}$ is not formed due to the absence of vacant d-orbital is not enough orbitals to accommodate extra electrons.
Hence, both asseration and reasons are correct and reason is the correct explanation of the assertion.

AIIMS-2015

Classification of Elements and Periodicity in Properties

89379 Lattice energy of an ionic compound depends upon

1 charge on the ion only

2 size of the ion only

3 charge on the ion and size of the ion

4 packing of its ions only

Explanation:

From the Born-Lande equation-
$U = \frac{N_{A} {Mz}^{+} z^{-} e^{2}}{4 π ϵ_{0} r} (1 - \frac{1}{n})$
Where, $U =$ Lattice energy of crystal
$N_{A} =$ Avogadro's number
$M =$ Molecular weight
$z^{+}, z^{-} =$ Charge of cation and anion
$r =$ Radius of the atom
$n =$ Exponent
From the above equation it is clear that lattice energy depends upon charge on the ion and size of the ion.

AP - EAMCET(MEDICAL) - 2009

Classification of Elements and Periodicity in Properties

89395 What is the correct relationship between the pH of isomolar solutions of sodium oxide, ${Na}_{2} O$ $({pH}_{1})$ , sodium sulphide, ${Na}_{2} S ({pH}_{2})$ , sodium selenide, ${Na}_{2} Se ({pH}_{3})$ and sodium telluride ${Na}_{2} T e ({pH}_{4})$ ?

1

{pH}_{1} > {pH}_{2} > {pH}_{3} > {pH}_{4}

2

{pH}_{1} > {pH}_{2} = {pH}_{3} > {pH}_{4}

3

{pH}_{1} > {pH}_{2} < {pH}_{3} < {pH}_{4}

4

{pH}_{1} > {pH}_{2} < {pH}_{3} \approx {pH}_{4}

Explanation:

The basicity of these oxides decreases, when placed in water, the basicity of solution also decreases.
${Na}_{2} O + 2 H_{2} O ⟶ 2 NaOH + H_{2} O$
${Na}_{2} S + 2 H_{2} O ⟶ 2 NaOH + H_{2} S$
${Na}_{2} Se + 2 H_{2} O ⟶ 2 NaOH + H_{2} Se$
${Na}_{2} Te + 2 H_{2} O ⟶ 2 NaOH + H_{2} Te$
Hence, the order of neutralisation of $NaOH$ is $H_{2} Te >$ $H_{2} Se > H_{2} S > H_{2} O$ .
Their aqueous solution have the following order of basic character due to neutralization of $NaOH$ with ${Na}_{2} O > {Na}_{2} S > {Na}_{2} Se > {Na}_{2} Te$ or $({pH}_{1}) > ({pH}_{2}) > ({pH}_{3}) > ({PH}_{4})$

NEET-2005

Classification of Elements and Periodicity in Properties

89401 Among the following elements that has the lowest first ionization potential is

1 nitrogen

2 oxygen

3 fluorine

4 neon

Explanation:

$_{7} N = 1 s^{2} 2 s^{2} 2 p^{3}$
$\begin{aligned} _{8} O = 1 s^{2} 2 s^{2} 2 p^{4} \\ _{9} F = 1 s^{2} 2 s^{2} 2 p^{5} \\ _{10} Ne = 1 s^{2} 2 s^{2} 2 p^{6} \end{aligned}$
Nitrogen has half-filled orbital and Ne has filled orbital it show stable configuration but Oxygen has less stable configuration and less nuclear charge.

A.P.EAMCET 1994

Classification of Elements and Periodicity in Properties

89368 Assertion: Both $Be$ and $Al$ can form complexes such as ${BeF}_{4}^{2 -}$ and ${AlF}_{6}^{3 -}$ respectively, ${BeF}_{6}^{3 -}$ is not formed.
Reason: In case of $B e$ , no vacant d-orbital's are present in its outermost shell.

1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

$Be$ and $Al$ can form complexes such as ${BeF}_{4}^{2 -}$ and ${AlF}_{6}^{3 -}$ respectively, ${BeF}_{6}^{3 -}$ is not formed due to the absence of vacant d-orbital is not enough orbitals to accommodate extra electrons.
Hence, both asseration and reasons are correct and reason is the correct explanation of the assertion.

AIIMS-2015

Classification of Elements and Periodicity in Properties

89379 Lattice energy of an ionic compound depends upon

1 charge on the ion only

2 size of the ion only

3 charge on the ion and size of the ion

4 packing of its ions only

Explanation:

From the Born-Lande equation-
$U = \frac{N_{A} {Mz}^{+} z^{-} e^{2}}{4 π ϵ_{0} r} (1 - \frac{1}{n})$
Where, $U =$ Lattice energy of crystal
$N_{A} =$ Avogadro's number
$M =$ Molecular weight
$z^{+}, z^{-} =$ Charge of cation and anion
$r =$ Radius of the atom
$n =$ Exponent
From the above equation it is clear that lattice energy depends upon charge on the ion and size of the ion.

AP - EAMCET(MEDICAL) - 2009

Classification of Elements and Periodicity in Properties

89395 What is the correct relationship between the pH of isomolar solutions of sodium oxide, ${Na}_{2} O$ $({pH}_{1})$ , sodium sulphide, ${Na}_{2} S ({pH}_{2})$ , sodium selenide, ${Na}_{2} Se ({pH}_{3})$ and sodium telluride ${Na}_{2} T e ({pH}_{4})$ ?

1

{pH}_{1} > {pH}_{2} > {pH}_{3} > {pH}_{4}

2

{pH}_{1} > {pH}_{2} = {pH}_{3} > {pH}_{4}

3

{pH}_{1} > {pH}_{2} < {pH}_{3} < {pH}_{4}

4

{pH}_{1} > {pH}_{2} < {pH}_{3} \approx {pH}_{4}

Explanation:

The basicity of these oxides decreases, when placed in water, the basicity of solution also decreases.
${Na}_{2} O + 2 H_{2} O ⟶ 2 NaOH + H_{2} O$
${Na}_{2} S + 2 H_{2} O ⟶ 2 NaOH + H_{2} S$
${Na}_{2} Se + 2 H_{2} O ⟶ 2 NaOH + H_{2} Se$
${Na}_{2} Te + 2 H_{2} O ⟶ 2 NaOH + H_{2} Te$
Hence, the order of neutralisation of $NaOH$ is $H_{2} Te >$ $H_{2} Se > H_{2} S > H_{2} O$ .
Their aqueous solution have the following order of basic character due to neutralization of $NaOH$ with ${Na}_{2} O > {Na}_{2} S > {Na}_{2} Se > {Na}_{2} Te$ or $({pH}_{1}) > ({pH}_{2}) > ({pH}_{3}) > ({PH}_{4})$

NEET-2005

Classification of Elements and Periodicity in Properties

89401 Among the following elements that has the lowest first ionization potential is

1 nitrogen

2 oxygen

3 fluorine

4 neon

Explanation:

$_{7} N = 1 s^{2} 2 s^{2} 2 p^{3}$
$\begin{aligned} _{8} O = 1 s^{2} 2 s^{2} 2 p^{4} \\ _{9} F = 1 s^{2} 2 s^{2} 2 p^{5} \\ _{10} Ne = 1 s^{2} 2 s^{2} 2 p^{6} \end{aligned}$
Nitrogen has half-filled orbital and Ne has filled orbital it show stable configuration but Oxygen has less stable configuration and less nuclear charge.

A.P.EAMCET 1994

Classification of Elements and Periodicity in Properties

89368 Assertion: Both $Be$ and $Al$ can form complexes such as ${BeF}_{4}^{2 -}$ and ${AlF}_{6}^{3 -}$ respectively, ${BeF}_{6}^{3 -}$ is not formed.
Reason: In case of $B e$ , no vacant d-orbital's are present in its outermost shell.

1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

$Be$ and $Al$ can form complexes such as ${BeF}_{4}^{2 -}$ and ${AlF}_{6}^{3 -}$ respectively, ${BeF}_{6}^{3 -}$ is not formed due to the absence of vacant d-orbital is not enough orbitals to accommodate extra electrons.
Hence, both asseration and reasons are correct and reason is the correct explanation of the assertion.

AIIMS-2015

Classification of Elements and Periodicity in Properties

89379 Lattice energy of an ionic compound depends upon

1 charge on the ion only

2 size of the ion only

3 charge on the ion and size of the ion

4 packing of its ions only

Explanation:

From the Born-Lande equation-
$U = \frac{N_{A} {Mz}^{+} z^{-} e^{2}}{4 π ϵ_{0} r} (1 - \frac{1}{n})$
Where, $U =$ Lattice energy of crystal
$N_{A} =$ Avogadro's number
$M =$ Molecular weight
$z^{+}, z^{-} =$ Charge of cation and anion
$r =$ Radius of the atom
$n =$ Exponent
From the above equation it is clear that lattice energy depends upon charge on the ion and size of the ion.

AP - EAMCET(MEDICAL) - 2009

Classification of Elements and Periodicity in Properties

89395 What is the correct relationship between the pH of isomolar solutions of sodium oxide, ${Na}_{2} O$ $({pH}_{1})$ , sodium sulphide, ${Na}_{2} S ({pH}_{2})$ , sodium selenide, ${Na}_{2} Se ({pH}_{3})$ and sodium telluride ${Na}_{2} T e ({pH}_{4})$ ?

1

{pH}_{1} > {pH}_{2} > {pH}_{3} > {pH}_{4}

2

{pH}_{1} > {pH}_{2} = {pH}_{3} > {pH}_{4}

3

{pH}_{1} > {pH}_{2} < {pH}_{3} < {pH}_{4}

4

{pH}_{1} > {pH}_{2} < {pH}_{3} \approx {pH}_{4}

Explanation:

The basicity of these oxides decreases, when placed in water, the basicity of solution also decreases.
${Na}_{2} O + 2 H_{2} O ⟶ 2 NaOH + H_{2} O$
${Na}_{2} S + 2 H_{2} O ⟶ 2 NaOH + H_{2} S$
${Na}_{2} Se + 2 H_{2} O ⟶ 2 NaOH + H_{2} Se$
${Na}_{2} Te + 2 H_{2} O ⟶ 2 NaOH + H_{2} Te$
Hence, the order of neutralisation of $NaOH$ is $H_{2} Te >$ $H_{2} Se > H_{2} S > H_{2} O$ .
Their aqueous solution have the following order of basic character due to neutralization of $NaOH$ with ${Na}_{2} O > {Na}_{2} S > {Na}_{2} Se > {Na}_{2} Te$ or $({pH}_{1}) > ({pH}_{2}) > ({pH}_{3}) > ({PH}_{4})$

NEET-2005

Classification of Elements and Periodicity in Properties

89401 Among the following elements that has the lowest first ionization potential is

1 nitrogen

2 oxygen

3 fluorine

4 neon

Explanation:

$_{7} N = 1 s^{2} 2 s^{2} 2 p^{3}$
$\begin{aligned} _{8} O = 1 s^{2} 2 s^{2} 2 p^{4} \\ _{9} F = 1 s^{2} 2 s^{2} 2 p^{5} \\ _{10} Ne = 1 s^{2} 2 s^{2} 2 p^{6} \end{aligned}$
Nitrogen has half-filled orbital and Ne has filled orbital it show stable configuration but Oxygen has less stable configuration and less nuclear charge.

A.P.EAMCET 1994

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