Draw a Lewis structure for methyl isocyanate, \(\mathrm{CH}_{3} \mathrm{NCO}\), showing all valence electrons. Predict all bond angles in this molecule and the hybridization of each \(\mathrm{C}, \mathrm{N}\), and \(\mathrm{O}\).

To determine the total number of valence electrons for methyl isocyanate \(\mathrm{(CH}_{3} \mathrm{NCO)}\), we first find the number of valence electrons for each atom: - C (carbon) has 4 valence electrons (group 14 in the periodic table). - H (hydrogen) has 1 valence electron (group 1 in the periodic table). - N (nitrogen) has 5 valence electrons (group 15 in the periodic table). - O (oxygen) has 6 valence electrons (group 16 in the periodic table). In methyl isocyanate, we have 1 carbon, 3 hydrogen, 1 nitrogen, and 1 oxygen atom. Therefore, the total number of valence electrons is: \(4 + 3(1) + 5 + 6 = 18\).

Now, we'll draw the Lewis structure for methyl isocyanate, remembering that the molecule's atomic arrangement is \(\mathrm{CH}_{3} \mathrm{NCO}\). Start by arranging the atoms and connecting them with single bonds, which use 2 valence electrons each. H - C - N - C (double bond) O Now, complete the octet for each atom by adding lone pairs: - Carbon in \(\mathrm{CH}_{3}\) group: already has 4 bonds (4 electrons). - Hydrogen atoms: Each has 1 bond (1 electron) and doesn't need lone pairs. - Nitrogen: has 3 bonds (6 electrons) so add 1 lone pair (2 electrons). - Carbon attached to N and O: has 4 bonds (8 electrons) with a double bond to the O. - Oxygen: has 2 bonds (4 electrons) from the double bond, add 2 lone pairs (4 electrons). Final Lewis Structure: H - C - N (lone pair) - C (double bond) O (2 lone pairs)

Using the VSEPR theory, we can predict the bond angles in the molecule: - \(\angle{\mathrm{H-C-H}}\): The C atom in the \(\mathrm{CH}_{3}\) group has 3 bonding and 0 lone pair electrons (AX3). It is trigonal planar with a bond angle of 120°. - \(\angle{\mathrm{C-N-C}}\): The N atom has 3 bonding and 1 lone pair electrons (AX3E). Its structure is trigonal pyramidal with a bond angle of around 107°. - \(\angle{\mathrm{N-C-O}}\): The C atom attached to N and O has 4 bonding and 0 lone pair electrons (AX4). Its structure is tetrahedral with a bond angle of 109.5°.

To find the hybridization of each C, N, and O atom, we count the number of electron groups (bonding + lone pairs) around each atom: - Carbon in the \(\mathrm{CH}_{3}\) group: 3 electron groups → sp2 hybridization. - Nitrogen: 4 electron groups → sp3 hybridization. - Carbon attached to N and O: 4 electron groups → sp3 hybridization. - Oxygen: 3 electron groups → sp2 hybridization. In summary, the hybridization of each C, N, and O atom in methyl isocyanate is as follows: - C in the \(\mathrm{CH}_{3}\) group: sp2 - N: sp3 - C attached to N and O: sp3 - O: sp2