238940
The correct electronic configuration and spinonly magnetic moment (BM) of $\mathrm{Gd}^{3+}(Z=64)$, respectively, are
1 $[\mathrm{Xe}] 4 \mathrm{f}^7$ and 7.9
2 $[\mathrm{Xe}] 5 \mathrm{f}^7$ and 7.9
3 $[\mathrm{Xe}] 5 \mathrm{f}^7$ and 8.9
4 $[\mathrm{Xe}] 4 \mathrm{f}^7$ and 8.9
Explanation:
Given, $[\mathrm{Gd}]^{3+}$ To find electronic configuration and magnetic moment of $[\mathrm{Gd}]^{3+}$ - Electronic configuration of $[\mathrm{Gd}]^{3+}:[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^0 6 \mathrm{~s}^0$ Hence number of unpaired electron $\mathrm{n}=7$ Hence magnetic moment $\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{63}=7.9 \mathrm{BM}$ Hence electronic configuration is $[\mathrm{Xe}] 4 \mathrm{f}^7$ Magnetic moment is $7.9 \mathrm{BM}$
(JEE Main 2020
Structure of Atom
238944
Which of the following atoms in its ground state has the highest number of unpaired electrons?
1 Chromium (24)
2 Iron (26)
3 Manganese (25)
4 Vanadium (23)
Explanation:
The electronic configuration of the give element is- Chromium (24): $\left(3 d^5 4 \mathrm{~s}^1\right)-6$ unpaired electron. Iron (26) : $\left(3 \mathrm{~d}^6 4 \mathrm{~s}^2\right)-4$ unpaired electron. Manganese (25) : $\left(3 \mathrm{~d}^5 4 \mathrm{~s}^2\right)-5$ unpaired electron. Vanadium (23) : $\left(3 \mathrm{~d}^3 3 \mathrm{~s}^2\right)-3$ unpaired electron. So, ${ }_{24} \mathrm{Cr}$ has the highest number of unpaired electrons.
J and K CET-(2016)
Structure of Atom
238946
Which electronic configuration will show the highest first ionization potential?
If $n=6$, then the correct sequence for filling of electrons will be $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow \mathrm{np}$. It will be $6 \mathrm{~s} \rightarrow 4 \mathrm{f} \rightarrow 5 \mathrm{~d} \rightarrow 6 \mathrm{p} \text {. }$ The energy of $4 \mathrm{f}$ and $5 \mathrm{~d}$ orbitals are in between the energies of $4 s$ and $6 p$ orbitals. Note:- Higher is the $(\mathrm{n}+\ell)$, higher will be the energy of the orbital. For $\mathrm{n}=6$ $\begin{aligned} & 6 s=6+0=6 \\ & 6 p=6+1=7 \\ & 5 d=5+2=7 \\ & 4 f=4+3=7 \end{aligned}$ Hence, the correct order is: $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow \mathrm{np}$
In general orbitals with lower energy are present near to the nucleus. This can be calculate by $(\mathrm{n}+1)$ rule. (i) Lower be the sum of $(n+1)$ value, nearest the orbital to the nucleus. (ii) And for the same sum, if the value of $\mathrm{n}$ will be lower, then its energy will also be low and it will be nearest to the nucleus. Therefore, from the given options- (a.) $4 \mathrm{f} \rightarrow(\ell=3) \therefore \mathrm{n}+\ell=4+3=7$ (b.) $5 \mathrm{~d} \rightarrow(\ell=2) \therefore \mathrm{n}+\ell=5+2=7$ (c.) $4 \mathrm{~s} \rightarrow(\ell=0) \therefore \mathrm{n}+\ell=4+0=4$ (d.) $7 \mathrm{p} \rightarrow(\ell=1) \therefore \mathrm{n}+\ell=7+1=8$
238940
The correct electronic configuration and spinonly magnetic moment (BM) of $\mathrm{Gd}^{3+}(Z=64)$, respectively, are
1 $[\mathrm{Xe}] 4 \mathrm{f}^7$ and 7.9
2 $[\mathrm{Xe}] 5 \mathrm{f}^7$ and 7.9
3 $[\mathrm{Xe}] 5 \mathrm{f}^7$ and 8.9
4 $[\mathrm{Xe}] 4 \mathrm{f}^7$ and 8.9
Explanation:
Given, $[\mathrm{Gd}]^{3+}$ To find electronic configuration and magnetic moment of $[\mathrm{Gd}]^{3+}$ - Electronic configuration of $[\mathrm{Gd}]^{3+}:[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^0 6 \mathrm{~s}^0$ Hence number of unpaired electron $\mathrm{n}=7$ Hence magnetic moment $\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{63}=7.9 \mathrm{BM}$ Hence electronic configuration is $[\mathrm{Xe}] 4 \mathrm{f}^7$ Magnetic moment is $7.9 \mathrm{BM}$
(JEE Main 2020
Structure of Atom
238944
Which of the following atoms in its ground state has the highest number of unpaired electrons?
1 Chromium (24)
2 Iron (26)
3 Manganese (25)
4 Vanadium (23)
Explanation:
The electronic configuration of the give element is- Chromium (24): $\left(3 d^5 4 \mathrm{~s}^1\right)-6$ unpaired electron. Iron (26) : $\left(3 \mathrm{~d}^6 4 \mathrm{~s}^2\right)-4$ unpaired electron. Manganese (25) : $\left(3 \mathrm{~d}^5 4 \mathrm{~s}^2\right)-5$ unpaired electron. Vanadium (23) : $\left(3 \mathrm{~d}^3 3 \mathrm{~s}^2\right)-3$ unpaired electron. So, ${ }_{24} \mathrm{Cr}$ has the highest number of unpaired electrons.
J and K CET-(2016)
Structure of Atom
238946
Which electronic configuration will show the highest first ionization potential?
If $n=6$, then the correct sequence for filling of electrons will be $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow \mathrm{np}$. It will be $6 \mathrm{~s} \rightarrow 4 \mathrm{f} \rightarrow 5 \mathrm{~d} \rightarrow 6 \mathrm{p} \text {. }$ The energy of $4 \mathrm{f}$ and $5 \mathrm{~d}$ orbitals are in between the energies of $4 s$ and $6 p$ orbitals. Note:- Higher is the $(\mathrm{n}+\ell)$, higher will be the energy of the orbital. For $\mathrm{n}=6$ $\begin{aligned} & 6 s=6+0=6 \\ & 6 p=6+1=7 \\ & 5 d=5+2=7 \\ & 4 f=4+3=7 \end{aligned}$ Hence, the correct order is: $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow \mathrm{np}$
In general orbitals with lower energy are present near to the nucleus. This can be calculate by $(\mathrm{n}+1)$ rule. (i) Lower be the sum of $(n+1)$ value, nearest the orbital to the nucleus. (ii) And for the same sum, if the value of $\mathrm{n}$ will be lower, then its energy will also be low and it will be nearest to the nucleus. Therefore, from the given options- (a.) $4 \mathrm{f} \rightarrow(\ell=3) \therefore \mathrm{n}+\ell=4+3=7$ (b.) $5 \mathrm{~d} \rightarrow(\ell=2) \therefore \mathrm{n}+\ell=5+2=7$ (c.) $4 \mathrm{~s} \rightarrow(\ell=0) \therefore \mathrm{n}+\ell=4+0=4$ (d.) $7 \mathrm{p} \rightarrow(\ell=1) \therefore \mathrm{n}+\ell=7+1=8$
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Structure of Atom
238940
The correct electronic configuration and spinonly magnetic moment (BM) of $\mathrm{Gd}^{3+}(Z=64)$, respectively, are
1 $[\mathrm{Xe}] 4 \mathrm{f}^7$ and 7.9
2 $[\mathrm{Xe}] 5 \mathrm{f}^7$ and 7.9
3 $[\mathrm{Xe}] 5 \mathrm{f}^7$ and 8.9
4 $[\mathrm{Xe}] 4 \mathrm{f}^7$ and 8.9
Explanation:
Given, $[\mathrm{Gd}]^{3+}$ To find electronic configuration and magnetic moment of $[\mathrm{Gd}]^{3+}$ - Electronic configuration of $[\mathrm{Gd}]^{3+}:[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^0 6 \mathrm{~s}^0$ Hence number of unpaired electron $\mathrm{n}=7$ Hence magnetic moment $\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{63}=7.9 \mathrm{BM}$ Hence electronic configuration is $[\mathrm{Xe}] 4 \mathrm{f}^7$ Magnetic moment is $7.9 \mathrm{BM}$
(JEE Main 2020
Structure of Atom
238944
Which of the following atoms in its ground state has the highest number of unpaired electrons?
1 Chromium (24)
2 Iron (26)
3 Manganese (25)
4 Vanadium (23)
Explanation:
The electronic configuration of the give element is- Chromium (24): $\left(3 d^5 4 \mathrm{~s}^1\right)-6$ unpaired electron. Iron (26) : $\left(3 \mathrm{~d}^6 4 \mathrm{~s}^2\right)-4$ unpaired electron. Manganese (25) : $\left(3 \mathrm{~d}^5 4 \mathrm{~s}^2\right)-5$ unpaired electron. Vanadium (23) : $\left(3 \mathrm{~d}^3 3 \mathrm{~s}^2\right)-3$ unpaired electron. So, ${ }_{24} \mathrm{Cr}$ has the highest number of unpaired electrons.
J and K CET-(2016)
Structure of Atom
238946
Which electronic configuration will show the highest first ionization potential?
If $n=6$, then the correct sequence for filling of electrons will be $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow \mathrm{np}$. It will be $6 \mathrm{~s} \rightarrow 4 \mathrm{f} \rightarrow 5 \mathrm{~d} \rightarrow 6 \mathrm{p} \text {. }$ The energy of $4 \mathrm{f}$ and $5 \mathrm{~d}$ orbitals are in between the energies of $4 s$ and $6 p$ orbitals. Note:- Higher is the $(\mathrm{n}+\ell)$, higher will be the energy of the orbital. For $\mathrm{n}=6$ $\begin{aligned} & 6 s=6+0=6 \\ & 6 p=6+1=7 \\ & 5 d=5+2=7 \\ & 4 f=4+3=7 \end{aligned}$ Hence, the correct order is: $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow \mathrm{np}$
In general orbitals with lower energy are present near to the nucleus. This can be calculate by $(\mathrm{n}+1)$ rule. (i) Lower be the sum of $(n+1)$ value, nearest the orbital to the nucleus. (ii) And for the same sum, if the value of $\mathrm{n}$ will be lower, then its energy will also be low and it will be nearest to the nucleus. Therefore, from the given options- (a.) $4 \mathrm{f} \rightarrow(\ell=3) \therefore \mathrm{n}+\ell=4+3=7$ (b.) $5 \mathrm{~d} \rightarrow(\ell=2) \therefore \mathrm{n}+\ell=5+2=7$ (c.) $4 \mathrm{~s} \rightarrow(\ell=0) \therefore \mathrm{n}+\ell=4+0=4$ (d.) $7 \mathrm{p} \rightarrow(\ell=1) \therefore \mathrm{n}+\ell=7+1=8$
238940
The correct electronic configuration and spinonly magnetic moment (BM) of $\mathrm{Gd}^{3+}(Z=64)$, respectively, are
1 $[\mathrm{Xe}] 4 \mathrm{f}^7$ and 7.9
2 $[\mathrm{Xe}] 5 \mathrm{f}^7$ and 7.9
3 $[\mathrm{Xe}] 5 \mathrm{f}^7$ and 8.9
4 $[\mathrm{Xe}] 4 \mathrm{f}^7$ and 8.9
Explanation:
Given, $[\mathrm{Gd}]^{3+}$ To find electronic configuration and magnetic moment of $[\mathrm{Gd}]^{3+}$ - Electronic configuration of $[\mathrm{Gd}]^{3+}:[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^0 6 \mathrm{~s}^0$ Hence number of unpaired electron $\mathrm{n}=7$ Hence magnetic moment $\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{63}=7.9 \mathrm{BM}$ Hence electronic configuration is $[\mathrm{Xe}] 4 \mathrm{f}^7$ Magnetic moment is $7.9 \mathrm{BM}$
(JEE Main 2020
Structure of Atom
238944
Which of the following atoms in its ground state has the highest number of unpaired electrons?
1 Chromium (24)
2 Iron (26)
3 Manganese (25)
4 Vanadium (23)
Explanation:
The electronic configuration of the give element is- Chromium (24): $\left(3 d^5 4 \mathrm{~s}^1\right)-6$ unpaired electron. Iron (26) : $\left(3 \mathrm{~d}^6 4 \mathrm{~s}^2\right)-4$ unpaired electron. Manganese (25) : $\left(3 \mathrm{~d}^5 4 \mathrm{~s}^2\right)-5$ unpaired electron. Vanadium (23) : $\left(3 \mathrm{~d}^3 3 \mathrm{~s}^2\right)-3$ unpaired electron. So, ${ }_{24} \mathrm{Cr}$ has the highest number of unpaired electrons.
J and K CET-(2016)
Structure of Atom
238946
Which electronic configuration will show the highest first ionization potential?
If $n=6$, then the correct sequence for filling of electrons will be $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow \mathrm{np}$. It will be $6 \mathrm{~s} \rightarrow 4 \mathrm{f} \rightarrow 5 \mathrm{~d} \rightarrow 6 \mathrm{p} \text {. }$ The energy of $4 \mathrm{f}$ and $5 \mathrm{~d}$ orbitals are in between the energies of $4 s$ and $6 p$ orbitals. Note:- Higher is the $(\mathrm{n}+\ell)$, higher will be the energy of the orbital. For $\mathrm{n}=6$ $\begin{aligned} & 6 s=6+0=6 \\ & 6 p=6+1=7 \\ & 5 d=5+2=7 \\ & 4 f=4+3=7 \end{aligned}$ Hence, the correct order is: $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow \mathrm{np}$
In general orbitals with lower energy are present near to the nucleus. This can be calculate by $(\mathrm{n}+1)$ rule. (i) Lower be the sum of $(n+1)$ value, nearest the orbital to the nucleus. (ii) And for the same sum, if the value of $\mathrm{n}$ will be lower, then its energy will also be low and it will be nearest to the nucleus. Therefore, from the given options- (a.) $4 \mathrm{f} \rightarrow(\ell=3) \therefore \mathrm{n}+\ell=4+3=7$ (b.) $5 \mathrm{~d} \rightarrow(\ell=2) \therefore \mathrm{n}+\ell=5+2=7$ (c.) $4 \mathrm{~s} \rightarrow(\ell=0) \therefore \mathrm{n}+\ell=4+0=4$ (d.) $7 \mathrm{p} \rightarrow(\ell=1) \therefore \mathrm{n}+\ell=7+1=8$
238940
The correct electronic configuration and spinonly magnetic moment (BM) of $\mathrm{Gd}^{3+}(Z=64)$, respectively, are
1 $[\mathrm{Xe}] 4 \mathrm{f}^7$ and 7.9
2 $[\mathrm{Xe}] 5 \mathrm{f}^7$ and 7.9
3 $[\mathrm{Xe}] 5 \mathrm{f}^7$ and 8.9
4 $[\mathrm{Xe}] 4 \mathrm{f}^7$ and 8.9
Explanation:
Given, $[\mathrm{Gd}]^{3+}$ To find electronic configuration and magnetic moment of $[\mathrm{Gd}]^{3+}$ - Electronic configuration of $[\mathrm{Gd}]^{3+}:[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^0 6 \mathrm{~s}^0$ Hence number of unpaired electron $\mathrm{n}=7$ Hence magnetic moment $\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{63}=7.9 \mathrm{BM}$ Hence electronic configuration is $[\mathrm{Xe}] 4 \mathrm{f}^7$ Magnetic moment is $7.9 \mathrm{BM}$
(JEE Main 2020
Structure of Atom
238944
Which of the following atoms in its ground state has the highest number of unpaired electrons?
1 Chromium (24)
2 Iron (26)
3 Manganese (25)
4 Vanadium (23)
Explanation:
The electronic configuration of the give element is- Chromium (24): $\left(3 d^5 4 \mathrm{~s}^1\right)-6$ unpaired electron. Iron (26) : $\left(3 \mathrm{~d}^6 4 \mathrm{~s}^2\right)-4$ unpaired electron. Manganese (25) : $\left(3 \mathrm{~d}^5 4 \mathrm{~s}^2\right)-5$ unpaired electron. Vanadium (23) : $\left(3 \mathrm{~d}^3 3 \mathrm{~s}^2\right)-3$ unpaired electron. So, ${ }_{24} \mathrm{Cr}$ has the highest number of unpaired electrons.
J and K CET-(2016)
Structure of Atom
238946
Which electronic configuration will show the highest first ionization potential?
If $n=6$, then the correct sequence for filling of electrons will be $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow \mathrm{np}$. It will be $6 \mathrm{~s} \rightarrow 4 \mathrm{f} \rightarrow 5 \mathrm{~d} \rightarrow 6 \mathrm{p} \text {. }$ The energy of $4 \mathrm{f}$ and $5 \mathrm{~d}$ orbitals are in between the energies of $4 s$ and $6 p$ orbitals. Note:- Higher is the $(\mathrm{n}+\ell)$, higher will be the energy of the orbital. For $\mathrm{n}=6$ $\begin{aligned} & 6 s=6+0=6 \\ & 6 p=6+1=7 \\ & 5 d=5+2=7 \\ & 4 f=4+3=7 \end{aligned}$ Hence, the correct order is: $\mathrm{ns} \rightarrow(\mathrm{n}-2) \mathrm{f} \rightarrow(\mathrm{n}-1) \mathrm{d} \rightarrow \mathrm{np}$
In general orbitals with lower energy are present near to the nucleus. This can be calculate by $(\mathrm{n}+1)$ rule. (i) Lower be the sum of $(n+1)$ value, nearest the orbital to the nucleus. (ii) And for the same sum, if the value of $\mathrm{n}$ will be lower, then its energy will also be low and it will be nearest to the nucleus. Therefore, from the given options- (a.) $4 \mathrm{f} \rightarrow(\ell=3) \therefore \mathrm{n}+\ell=4+3=7$ (b.) $5 \mathrm{~d} \rightarrow(\ell=2) \therefore \mathrm{n}+\ell=5+2=7$ (c.) $4 \mathrm{~s} \rightarrow(\ell=0) \therefore \mathrm{n}+\ell=4+0=4$ (d.) $7 \mathrm{p} \rightarrow(\ell=1) \therefore \mathrm{n}+\ell=7+1=8$