Here is Exercise I.2 of R. Friedman's Algebraic Surfaces and Holomorphic Vector Bundles.
Let $X$ be a smooth surface in $\mathbb{P}^3$ of degree $d$. Show using standard facts about the cohomology of $\mathbb{P}^3$ that $q(X)=0$. ... Using the fact that $K_X=\mathcal{O}_X(d-4)$ by adjunction and the fact that $q(X)=0$, determine $c_1^2(X)$ and $p_g(X)$. Apply Noether's formula to find $b_2(X)$ and thus $h^{1,1}(X)$. ...... When is the intersection form on $X$ of Type I (odd)?
I am stuck at the last question about whether the intersection form is even or odd. I now only get a sufficient condition for Type I. I get $$c_1(X)^2=(d-4)^2d$$ and then $d$ is odd $\implies$ Type I.
I wonder if more can be said. It seems $d$ is the only variable here. Will the type be determined by $d$?
Yes, it is also true that $d$ is even implies that the intersection form is even.
This follows from the Wu formula $$ Q(w_2(TX), \alpha)=Q(\alpha, \alpha)\mod 2, $$ where $\alpha\in H^2(X, {\mathbb Z}/2)$, $Q$ is the intersection form, and $w_2(TX)\in H^2(X, {\mathbb Z}/2)$ is the second Stiefel-Whitney class.
Donaldson and Kronheimer's "Geometry of four-manifolds" has some explanation of this in (1.1.5). Also see Property of second Stiefel-Whitney class?
Since $w_2(TX)\equiv c_1(TX)\mod 2$, and $c_1(TX)=(4-d)h|_X$, where $h\in H^2({\mathbb C}P^3, {\mathbb Z})$ is the hyperplane class, we see that $w_2(TX)=0$ if $d$ is even. Then the intersection form is even. (This is also the case where the surface $X$ is spin.)