orbital is : (1) 000) (2) 2 (3) 3 (1) (UI H-N....N N In hydrogen azide (above) the bond orders of bonds (1) and ( man 18. (JEE(Main) 2018 Online (15-04-18), 411

Draw the molecuar orbital diagram of N_2 and calculate the bond order.

Using the data Delta_(f) H^(@) (Nf_(3), g) = - 114 kH mol^(-1) , Delta_(N -= N) H^(@) = 946 kJ mol^(-1) , and Delta_(f - f) H^(@) = 158 kJ mol^(-1) , calculate the average bond enthalpy of N - F bond in NF_(3) . Strategy : First write the thermochemical equation corresponding to Delta_(f) H^(@) (NF_(3), g) : (1)/(2) N_(2) (g) + (3)/(2) F_(2) (g) rarr NF_(3) (g) Now define Delta_(r) H^(@) of this reaction in terms of bonds made and bonds broken. Notice the 1//2 mol of N -= N bonds and 3//2 mol of F - F bond are broken, while 3 mole of N - F bonds are formed each NF_(3) has three N - F bonds: we are given the blood enthalpies of N -= N and F - F bonds, while the average bond enthalpy of N -F bond is not known. Applying Eq. for the reaction, we can calculate this unknown.

The hydrogen bonding ability of the isomeric 1^(@),2^(@) and 3^(@) alcohols decreases in the order

The enthalpy change (DeltaH) for the process, N_(2)H_(4)(g)to 2N(g)+4H(g) is is 1724 kJ mol^(-1) . If the bond energy of N-H bond in ammonia is 391 kJ mol^(-1) , what is the bond energy for N-N bond in N_(2)H_(4) ?

The enthalpy change (DeltaH) for the process, N_(2)H_(4)(g)to 2N(g)+4H(g) is is 1724 kJ mol^(-1) . If the bond energy of N-H bond in ammonia is 391 kJ mol^(-1) , what is the bond energy for N-N bond in N_(2)H_(4) ?

Give the Molecular Orbital Diagram of N_2 . Calculate the bond order of N_2 .