Chromium in chromyl chloride (CrO$_2$Cl$_2$):
Oxidation state of Cr is +6.
Electronic configuration: Cr$^{6+}$ = [Ar] 3d$^0$ 4s$^0$, meaning there are zero d-electrons.
Manganese in Mn (VII):
Oxidation state of Mn is +7.
Electronic configuration: Mn$^{7+}$ = [Ar] 3d$^0$ 4s$^0$, also zero d-electrons.
Thus, the number of d-electrons is the same for Cr in CrO$_2$Cl$_2$ and Mn in Mn (VII).
Given below are the quantum numbers for 4 electrons.
A. n=3, l=2, ml=1,ms=+\(\frac{1}{2}\)
B. n=4, l=1, ml=0,ms=+\(\frac{1}{2}\)
C. n=4, l=2, ml=–2,ms=–\(\frac{1}{2}\)
D. n=3, l=1, ml=–1,ms=+\(\frac{1}{2}\)
The correct order of increasing energy is
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32