Non-polar molecules have a symmetrical arrangement of atoms that results in no net dipole moment. In the given list: - \(CO_2\) is linear and symmetrical, so it is non-polar. - \(H_2\) is diatomic and non-polar because it is composed of identical atoms. - \(CH_4\) has a tetrahedral geometry with symmetrical bond distribution, making it non-polar. - \(BF_3\) has a trigonal planar geometry, which is symmetrical and therefore non-polar.
Other molecules like HF, \(H_2O\), \(SO_2\), \(NH_3\), HCl, and \(CHCl_3\) are polar due to their asymmetrical shapes or differences in electronegativity. Therefore, there are four non-polar molecules in the list.
The Correct answer is: 4
Identify the correct orders against the property mentioned:
A. H$_2$O $>$ NH$_3$ $>$ CHCl$_3$ - dipole moment
B. XeF$_4$ $>$ XeO$_3$ $>$ XeF$_2$ - number of lone pairs on central atom
C. O–H $>$ C–H $>$ N–O - bond length
D. N$_2$>O$_2$>H$_2$ - bond enthalpy
Choose the correct answer from the options given below:
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: