The problem requires identifying non-polar molecules from a given list. A molecule is non-polar when its net dipole moment is zero, often due to a symmetrical arrangement of polar bonds that cancel each other out.
Counting non-polar molecules, we have H2, CO2, CH4, and BF3. Hence, there are 4 non-polar molecules.
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
From the given following (A to D) cyclic structures, those which will not react with Tollen's reagent are : 
Compound 'P' undergoes the following sequence of reactions : (i) NH₃ (ii) $\Delta$ $\rightarrow$ Q (i) KOH, Br₂ (ii) CHCl₃, KOH (alc), $\Delta$ $\rightarrow$ NC-CH₃. 'P' is : 

Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
Consider the following two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.
Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to