ChemistryQ1–2 of 2 questions
Page 1 of 1 · English
1
ChemistryEasyMCQAIIMS · 1989
Which one is the electronic configuration of $Fe^{2+}$?
A
$1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^6$
B
$1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^4, 4s^2$
C
$1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^5, 4s^1$
D
None of these
Solution
(A) The atomic number of iron $(Fe)$ is $26$.
The ground state electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$ or $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^6 4s^2$.
When $Fe$ forms the $Fe^{2+}$ ion,it loses two electrons from the outermost shell,which is the $4s$ orbital.
Therefore,the electronic configuration of $Fe^{2+}$ becomes $[Ar] 3d^6$ or $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^6$.
The ground state electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$ or $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^6 4s^2$.
When $Fe$ forms the $Fe^{2+}$ ion,it loses two electrons from the outermost shell,which is the $4s$ orbital.
Therefore,the electronic configuration of $Fe^{2+}$ becomes $[Ar] 3d^6$ or $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^6$.
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2
ChemistryMCQAIIMS · 1989
In an $NPN$ transistor circuit,the collector current is $10\, mA$. If $90\%$ of the electrons emitted reach the collector,the emitter current $(i_E)$ and base current $(i_B)$ are given by
A
$i_E = -1\, mA, i_B = 9\, mA$
B
$i_E = 9\, mA, i_B = -1\, mA$
C
$i_E = 1\, mA, i_B = 11\, mA$
D
$i_E = 11\, mA, i_B = 1\, mA$
Solution
(D) Given that the collector current $I_C = 10\, mA$.
According to the problem,$90\%$ of the electrons emitted reach the collector,which means the collector current is $90\%$ of the emitter current:
$I_C = 0.90 \times I_E$
Substituting the value of $I_C$:
$10\, mA = 0.90 \times I_E$
$I_E = \frac{10}{0.9} \approx 11.11\, mA \approx 11\, mA$.
Using the relation $I_E = I_B + I_C$,we can find the base current:
$I_B = I_E - I_C$
$I_B = 11\, mA - 10\, mA = 1\, mA$.
Thus,$I_E = 11\, mA$ and $I_B = 1\, mA$.
According to the problem,$90\%$ of the electrons emitted reach the collector,which means the collector current is $90\%$ of the emitter current:
$I_C = 0.90 \times I_E$
Substituting the value of $I_C$:
$10\, mA = 0.90 \times I_E$
$I_E = \frac{10}{0.9} \approx 11.11\, mA \approx 11\, mA$.
Using the relation $I_E = I_B + I_C$,we can find the base current:
$I_B = I_E - I_C$
$I_B = 11\, mA - 10\, mA = 1\, mA$.
Thus,$I_E = 11\, mA$ and $I_B = 1\, mA$.
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