= Solution
The light states have the counting $3+4+1=8$. This suggests approximate <chiral symmetry breaking>
$$
\boxed{SU(3)_L\times SU(3)_R\longrightarrow SU(3)_V,\qquad 2(3^2-1)-(3^2-1)=8,}
$$
with a nonzero <quark chiral condensate> $\langle\bar q_R^{\,j}q_L^{\,i}\rangle\propto\delta^{ij}$ for the three light flavours $u,d,s$. The three <pions>, four <kaons> and predominantly octet <eta meson> are the resulting <pseudo-Goldstone bosons>. Their small nonzero <masses> reflect explicit <chiral symmetry> breaking by the light-quark <masses>; for example, $m_\pi^2$ is proportional to $m_u+m_d$ at leading order. The larger strange-quark <mass> makes the <kaons> and octet <eta meson> heavier than the <pions>, while <electromagnetic interactions> help split charged and neutral partners.<Gauge Theory by David Tong>[Chapter 5, chiral symmetry breaking and section 5.3 on adding quark masses]{ref}
Although the massless strong-interaction terms admit independent rotations of all six flavours, the charm, bottom and top <masses> are too large to treat six-flavour <chiral symmetry> as a small explicit breaking at the light-hadron scale. Full approximate $SU(6)_L\times SU(6)_R$ breaking would suggest 35 light modes, rather than this octet. The classical axial $U(1)$ is also broken by the <chiral anomaly>, so it does not supply a ninth equally light <Goldstone boson>.
\b[Terminology qualification:] these mesons have <spin> zero but negative <parity>: they are <pseudoscalar mesons>, with $J^P=0^-$, rather than positive-parity <scalar> <particles> with $J^P=0^+$. Calling the listed states “<scalar>” is acceptable only if it is being used loosely to mean <spin> zero.
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