QBManager

Structure of Atom

239025 The number of angular and radial nodes of $4 d$ orbital respectively are

1 3,1

2 1,2

3 3,0

4 2,1

Explanation:

: Number of radial node $= n - l - 1$
Where, $n =$ Principle quantum number
$l =$ Azimuthal quantum number
$∴$ For $4 d$
$n = 4, l = 2$
Radial node $= 4 - 2 - 1 = 1$
Angular node $= l = 2$
Thus, the number of angular and radial node of $4 d$ orbital is 2,1 respectively.

AP-EAMCET (Engg.) - 2014

Structure of Atom

239028 The number of radial nodes of $3 s$ and $2 p$ orbitals respectively are

1 0,2

2 2,0

3 1,2

4 2,1

Explanation:

Number of radial nodes $= n - ℓ - 1$
Where $n =$ no. of principle quantum number. $ℓ =$ Azimuthal quantum number.
For $3 s$
For $2 p$ $n = 3, ℓ = 0$
$\begin{array}{cc} n = 2, ℓ = 1 \\ Radial nodes = 3 - 0 - 1 & Radial nodes = 2 - 1 - 1 \end{array}$ $= 2 = 0$
The number of radial nodes for $3 s$ and $2 p$ are 2,0 respectively.

AP-EAMCET (Engg.) 2013

Structure of Atom

239038 The ion that is isoelectronic with $CO$ is

1

O_{2}^{-}

2

N_{2}^{+}

3

{CN}^{-}

4

O_{2}^{+}

Explanation:

: Number of electrons in
$\begin{aligned} CO = 6 + 8 = 14 \\ {CN}^{-} = 6 + 7 + 1 = 14 \\ N_{2}^{+} = 7 + 7 - 1 = 13 \\ N_{2}^{2 -} = 7 + 7 + 2 = 16 \\ NO = 7 + 8 + 1 = 16 \end{aligned}$
$CO$ is isoelectronic with ${CN}^{-}$ because both species have same number of electrons.
Isoelcetronic species refers to the species which have same number of electrons.
${CN}^{- 1}$ is isoelectronic with $CO$ , both loose 14 electron.

Assam CEE-2018

Structure of Atom

239055 The quantum number of four electrons are given below:
I. $n = 4, l = 2, m_{1} = - 2, m_{s} = - \frac{1}{2}$
II. $n = 3, l = 2, m_{1} = 1, m_{5} = + \frac{1}{2}$
III. $n = 4, l = 1, m_{1} = 0, m_{s} = + \frac{1}{2}$
IV. $n = 3, l = 1, m_{1} = 1, m_{s} = - \frac{1}{2}$
The correct order of their increasing energies will be

1 IV

<

III

<

II

< 1

2 I

<

II

<

III

<

IV

3 IV

<

II

<

III

<

I

4 I

<

III

<

II

<

IV

Explanation:

: Smaller the value of $(n + l)$ , smaller the energy if two or more sub-orbits have same values of $(n + l)$ suborbits with lower values of $n$ has lower energy. The $(n + l)$ value of the given option are as follows:
(I) $n = 4, l = 2; n + l = 6$
(II) $n = 3, l = 2; n + l = 5$
(III) $n = 4, l = 1; n + l = 5$
(IV) $n = 3, l = 1; n + l = 4$
Among (II) and (III) $n = 3$ has lower energy. Thus the correct order of their increasing energies will be IV $<$ II $<$ III $<$ I

[JEE Main 2019

Structure of Atom

239025 The number of angular and radial nodes of $4 d$ orbital respectively are

1 3,1

2 1,2

3 3,0

4 2,1

Explanation:

: Number of radial node $= n - l - 1$
Where, $n =$ Principle quantum number
$l =$ Azimuthal quantum number
$∴$ For $4 d$
$n = 4, l = 2$
Radial node $= 4 - 2 - 1 = 1$
Angular node $= l = 2$
Thus, the number of angular and radial node of $4 d$ orbital is 2,1 respectively.

AP-EAMCET (Engg.) - 2014

Structure of Atom

239028 The number of radial nodes of $3 s$ and $2 p$ orbitals respectively are

1 0,2

2 2,0

3 1,2

4 2,1

Explanation:

Number of radial nodes $= n - ℓ - 1$
Where $n =$ no. of principle quantum number. $ℓ =$ Azimuthal quantum number.
For $3 s$
For $2 p$ $n = 3, ℓ = 0$
$\begin{array}{cc} n = 2, ℓ = 1 \\ Radial nodes = 3 - 0 - 1 & Radial nodes = 2 - 1 - 1 \end{array}$ $= 2 = 0$
The number of radial nodes for $3 s$ and $2 p$ are 2,0 respectively.

AP-EAMCET (Engg.) 2013

Structure of Atom

239038 The ion that is isoelectronic with $CO$ is

1

O_{2}^{-}

2

N_{2}^{+}

3

{CN}^{-}

4

O_{2}^{+}

Explanation:

: Number of electrons in
$\begin{aligned} CO = 6 + 8 = 14 \\ {CN}^{-} = 6 + 7 + 1 = 14 \\ N_{2}^{+} = 7 + 7 - 1 = 13 \\ N_{2}^{2 -} = 7 + 7 + 2 = 16 \\ NO = 7 + 8 + 1 = 16 \end{aligned}$
$CO$ is isoelectronic with ${CN}^{-}$ because both species have same number of electrons.
Isoelcetronic species refers to the species which have same number of electrons.
${CN}^{- 1}$ is isoelectronic with $CO$ , both loose 14 electron.

Assam CEE-2018

Structure of Atom

239055 The quantum number of four electrons are given below:
I. $n = 4, l = 2, m_{1} = - 2, m_{s} = - \frac{1}{2}$
II. $n = 3, l = 2, m_{1} = 1, m_{5} = + \frac{1}{2}$
III. $n = 4, l = 1, m_{1} = 0, m_{s} = + \frac{1}{2}$
IV. $n = 3, l = 1, m_{1} = 1, m_{s} = - \frac{1}{2}$
The correct order of their increasing energies will be

1 IV

<

III

<

II

< 1

2 I

<

II

<

III

<

IV

3 IV

<

II

<

III

<

I

4 I

<

III

<

II

<

IV

Explanation:

: Smaller the value of $(n + l)$ , smaller the energy if two or more sub-orbits have same values of $(n + l)$ suborbits with lower values of $n$ has lower energy. The $(n + l)$ value of the given option are as follows:
(I) $n = 4, l = 2; n + l = 6$
(II) $n = 3, l = 2; n + l = 5$
(III) $n = 4, l = 1; n + l = 5$
(IV) $n = 3, l = 1; n + l = 4$
Among (II) and (III) $n = 3$ has lower energy. Thus the correct order of their increasing energies will be IV $<$ II $<$ III $<$ I

[JEE Main 2019

Structure of Atom

239025 The number of angular and radial nodes of $4 d$ orbital respectively are

1 3,1

2 1,2

3 3,0

4 2,1

Explanation:

: Number of radial node $= n - l - 1$
Where, $n =$ Principle quantum number
$l =$ Azimuthal quantum number
$∴$ For $4 d$
$n = 4, l = 2$
Radial node $= 4 - 2 - 1 = 1$
Angular node $= l = 2$
Thus, the number of angular and radial node of $4 d$ orbital is 2,1 respectively.

AP-EAMCET (Engg.) - 2014

Structure of Atom

239028 The number of radial nodes of $3 s$ and $2 p$ orbitals respectively are

1 0,2

2 2,0

3 1,2

4 2,1

Explanation:

Number of radial nodes $= n - ℓ - 1$
Where $n =$ no. of principle quantum number. $ℓ =$ Azimuthal quantum number.
For $3 s$
For $2 p$ $n = 3, ℓ = 0$
$\begin{array}{cc} n = 2, ℓ = 1 \\ Radial nodes = 3 - 0 - 1 & Radial nodes = 2 - 1 - 1 \end{array}$ $= 2 = 0$
The number of radial nodes for $3 s$ and $2 p$ are 2,0 respectively.

AP-EAMCET (Engg.) 2013

Structure of Atom

239038 The ion that is isoelectronic with $CO$ is

1

O_{2}^{-}

2

N_{2}^{+}

3

{CN}^{-}

4

O_{2}^{+}

Explanation:

: Number of electrons in
$\begin{aligned} CO = 6 + 8 = 14 \\ {CN}^{-} = 6 + 7 + 1 = 14 \\ N_{2}^{+} = 7 + 7 - 1 = 13 \\ N_{2}^{2 -} = 7 + 7 + 2 = 16 \\ NO = 7 + 8 + 1 = 16 \end{aligned}$
$CO$ is isoelectronic with ${CN}^{-}$ because both species have same number of electrons.
Isoelcetronic species refers to the species which have same number of electrons.
${CN}^{- 1}$ is isoelectronic with $CO$ , both loose 14 electron.

Assam CEE-2018

Structure of Atom

239055 The quantum number of four electrons are given below:
I. $n = 4, l = 2, m_{1} = - 2, m_{s} = - \frac{1}{2}$
II. $n = 3, l = 2, m_{1} = 1, m_{5} = + \frac{1}{2}$
III. $n = 4, l = 1, m_{1} = 0, m_{s} = + \frac{1}{2}$
IV. $n = 3, l = 1, m_{1} = 1, m_{s} = - \frac{1}{2}$
The correct order of their increasing energies will be

1 IV

<

III

<

II

< 1

2 I

<

II

<

III

<

IV

3 IV

<

II

<

III

<

I

4 I

<

III

<

II

<

IV

Explanation:

: Smaller the value of $(n + l)$ , smaller the energy if two or more sub-orbits have same values of $(n + l)$ suborbits with lower values of $n$ has lower energy. The $(n + l)$ value of the given option are as follows:
(I) $n = 4, l = 2; n + l = 6$
(II) $n = 3, l = 2; n + l = 5$
(III) $n = 4, l = 1; n + l = 5$
(IV) $n = 3, l = 1; n + l = 4$
Among (II) and (III) $n = 3$ has lower energy. Thus the correct order of their increasing energies will be IV $<$ II $<$ III $<$ I

[JEE Main 2019

Structure of Atom

239025 The number of angular and radial nodes of $4 d$ orbital respectively are

1 3,1

2 1,2

3 3,0

4 2,1

Explanation:

: Number of radial node $= n - l - 1$
Where, $n =$ Principle quantum number
$l =$ Azimuthal quantum number
$∴$ For $4 d$
$n = 4, l = 2$
Radial node $= 4 - 2 - 1 = 1$
Angular node $= l = 2$
Thus, the number of angular and radial node of $4 d$ orbital is 2,1 respectively.

AP-EAMCET (Engg.) - 2014

Structure of Atom

239028 The number of radial nodes of $3 s$ and $2 p$ orbitals respectively are

1 0,2

2 2,0

3 1,2

4 2,1

Explanation:

Number of radial nodes $= n - ℓ - 1$
Where $n =$ no. of principle quantum number. $ℓ =$ Azimuthal quantum number.
For $3 s$
For $2 p$ $n = 3, ℓ = 0$
$\begin{array}{cc} n = 2, ℓ = 1 \\ Radial nodes = 3 - 0 - 1 & Radial nodes = 2 - 1 - 1 \end{array}$ $= 2 = 0$
The number of radial nodes for $3 s$ and $2 p$ are 2,0 respectively.

AP-EAMCET (Engg.) 2013

Structure of Atom

239038 The ion that is isoelectronic with $CO$ is

1

O_{2}^{-}

2

N_{2}^{+}

3

{CN}^{-}

4

O_{2}^{+}

Explanation:

: Number of electrons in
$\begin{aligned} CO = 6 + 8 = 14 \\ {CN}^{-} = 6 + 7 + 1 = 14 \\ N_{2}^{+} = 7 + 7 - 1 = 13 \\ N_{2}^{2 -} = 7 + 7 + 2 = 16 \\ NO = 7 + 8 + 1 = 16 \end{aligned}$
$CO$ is isoelectronic with ${CN}^{-}$ because both species have same number of electrons.
Isoelcetronic species refers to the species which have same number of electrons.
${CN}^{- 1}$ is isoelectronic with $CO$ , both loose 14 electron.

Assam CEE-2018

Structure of Atom

239055 The quantum number of four electrons are given below:
I. $n = 4, l = 2, m_{1} = - 2, m_{s} = - \frac{1}{2}$
II. $n = 3, l = 2, m_{1} = 1, m_{5} = + \frac{1}{2}$
III. $n = 4, l = 1, m_{1} = 0, m_{s} = + \frac{1}{2}$
IV. $n = 3, l = 1, m_{1} = 1, m_{s} = - \frac{1}{2}$
The correct order of their increasing energies will be

1 IV

<

III

<

II

< 1

2 I

<

II

<

III

<

IV

3 IV

<

II

<

III

<

I

4 I

<

III

<

II

<

IV

Explanation:

: Smaller the value of $(n + l)$ , smaller the energy if two or more sub-orbits have same values of $(n + l)$ suborbits with lower values of $n$ has lower energy. The $(n + l)$ value of the given option are as follows:
(I) $n = 4, l = 2; n + l = 6$
(II) $n = 3, l = 2; n + l = 5$
(III) $n = 4, l = 1; n + l = 5$
(IV) $n = 3, l = 1; n + l = 4$
Among (II) and (III) $n = 3$ has lower energy. Thus the correct order of their increasing energies will be IV $<$ II $<$ III $<$ I

[JEE Main 2019