Nuclear magnetic shieldings of both carbon and hydrogen atoms of haluro methyl molecules are highly influenced by the substitution of one or more hydrogen by halogen heavy atoms. We applied the linear response elimination of small components, LRESC, formalism to calculate such shieldings and learn whether including only few terms is enough for getting quantitative reproduction of the total shieldings or not. First, we discuss the contribution of all leading relativistic corrections to σ(C), in CHX(2)I molecular models with X = H, F, and Cl, and show that spin-orbit (SO) effects are the main ones. After adding the SO effects to the non-relativistic (NR) results, we obtain ~ 97% (93%) of the total LRESC values for σ(C) (σ(H)). The magnitude of SO terms increases when the halogen atom becomes heavier. In this case, such contributions to σ(C) can be extrapolated as a function of Z, the halogen atomic number. Furthermore, when paramagnetic spin-orbit (PSO) contributions are also considered, we obtain results that are within 1% of the total LRESC value. Then we study in detail the main electronic mechanisms involved to contribute C and H shieldings on CH(n)X(4 - n) (n = 1, 3), and CHXYZ (X, Y, Z = F, Cl, Br, I) model compounds. The pattern of σ(C) for all series of compounds follows a normal halogen dependence (NHD), though with different rate of increase. A special family of compounds is that of CHF(2)X for which σ(nr)(C) follows an inverse halogen dependence though the total shielding have a NHD due to the SO contributions. For the series CH(3)X (X = F, Cl, Br and I), we found that σ(SO) ~ Z(X) (2.53). Another important finding of this work is the logarithmic dependence of σ(SO)(C) with the substituent atomic number: ln σ(SO)(C) = A(X) + a(X) Z(Y) for both family of compounds CH(2)XY and CHX(2)Y. We also performed four-component calculations using the spin-free Hamiltonian to obtain SO contributions within a four-component framework.